Tutors Answer Your Questions about Probabilityandstatistics (FREE)
Question 994671: a soldier hits her target with a probability of 3/10. How many times should she shoot at the target to ensure at least 70% chance of hitting the target?
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! You want 0.7, the probability of missing, to be raised to the x power to where it is less than 0.3, for at that point, there will be a greater than 70% chance of hitting the target.
0.7^x=0.3
ln both sides
x ln(.7)=ln(.3)
0.357x= 1.204
x is greater than 3 but less than 4. Therefore, she has to shoot 4 times.
0.7^4=0.2401, and the probability of 4 misses in a row is 76%, and the probability of hitting the target is at least 70%.
Question 994679: if 7 mangoes are chosen at random from 10 mangoes of which 3 are rotten, what is the probability that exactly one is not rotten and the probability that at least one is rotten?
Found 2 solutions by stanbon, Boreal: Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! if 7 mangoes are chosen at random from 10 mangoes of which 3 are rotten, what is the probability that exactly one is not rotten

Binomial Problem with n = 7 and p(rotten) = 3/10; p(not rotten) = 7/10

P(x = 1 not rotten) = 7C1*(7/10)*(3/10)^6 = binompdf(7,7/10,1) = 0.0036

and the probability that at least one is rotten?
P(1<= x <= 7) = 1  P(x = 0 rotten) = 1  (7/10)^7 = 0.9177

Cheers,
Stan H.
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! probability of one being not rotten is 0.7.
Exactly 1 is 7(0.7)^1(0.3)^6, the 7 being the number of ways it can be chosen, first, second, third, fourth, fifth, sixth, or seventh.
=0.00357
=========
for at least one is rotten: Probability none is rotten and subtract that from 1. That is easier than calculating 1,2,3,4,5,6,7.
The probability that none is rotten is 0.7^7=0.0824
The probability that at least one is rotten is the complement, or 0.918
Question 994676: in one college the second year mathematics students in mathematics have a 50% chance of passing their mathematics examination. Of the 18 second year mathematics students this year, calculate exactly the probability that they all pass their examination and that exactly one student fails and hence calculate the exactly probability that no more 2 students fail.
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! This is a headtails problem
(0.5)^18=3.815 x 10^(6)
=======
That 1 person fails is 18(0.5)^1 (0.5)^17, the 18 being the 18 different ways this can happen.
=6.866 X 10(5)
=======
probability 2 fail equals (18C2)(0.5)^2*(0.5)^16=
=153(0.5^18)=5.798 x 10^(4)
The probability that no more than two fail is the sum of the above, or
0.000003815
0.00006866
0.0005798
=0.0006523
Question 994662: the sex distribution being equally probable, compute the expected number of girls in a family with 10 children and what is the probability that the expected number of girls occurs?
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! the sex distribution being equally probable, compute the expected number of girls in a family with 10 children and what is the probability that the expected number of girls occurs?

P(girl on each birth) = 1/2

E(X when X = 10) = (1/2)10 = 5

P(x = 5) = 10C5(1/2)^10 = 0.24609..
Cheers,
Stan H.
Question 994555: suppose the probability remain 20% but we take sample from 100 archers.
what is the standard error of proportion?
i have it set 100(.20)=20 100 (1.20)=80 SQUARE ROOT ALL THIS TOGETHER >>
.20(120)/100 BUT I AM GETTING A WEIRD ANSWER.
THANKS
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
You have posted half (or perhaps less) of a problem here. And you have left your "Caps Lock" on again. All caps is the text communications equivalent of shouting. Shouting is both annoying and rude, so stop it, please.
John
My calculator said it, I believe it, that settles it
Question 994551: THE AVERAGE TIME IT TAKES SOMEONE TO FINISH THE 2.4 MILE SWIM AT IRONMAN LOUISVILLE IS APPROXIMATELY 86 MIN WITH STANDARD DEVIATION OF 10 MIN. ASSUME THE DISTRIBUTION IS APPROX NORMAL.
WHAT IS PROBABILITY THAT A RANDOMLY SELECTED WILL SWIM UNDER 60 MINUTES?
I DO NOT KNOW WHAT FORMULA TO USE FOR THIS.. MY GUESS WAS XMEAN/STAND DEVIATION
SOMETHING LIKE P(X<60) 8660/10=2.6 BUT I HAVE FEELING I AM DOING IT WRONG. CAN YOU PLEASE SHOW ME IN STEPS?
THANKS
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
Sorry, I can't understand you when you shout. Turn off your "Caps Lock" and repost. Thank you.
John
My calculator said it, I believe it, that settles it
Question 994434: A sample of 100 wood and 100 graphite tennis rackets are taken from the warehouse. If 9 wood
and 19 graphite are defective and one racket is randomly selected from the sample, find the probability that the racket is wood or defective.
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! Or means either one.
Table
Wood ;;;;;;;;;;; Graphite ;;; Total
Defective;;;;;; 9 ;;;;;;;;;;; 19;;; ;;28
nonD;;;;;; 91 ;;;;;;;;;;;;;;; 81;;;;;;;172
Total100100200
Probability Wood is (100/200)
Probability defective is (28/200)
Probability BOTH is (9/200)
The answer is the sum of the first two minus the third, since we have double counted.
It is (100+289)/200=(119/200). It is the everything but graphite and nondefective.
Question 994345: A basket contains 8 oranges, 3 mangoes and 9 apples. If 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! A basket contains 8 oranges, 3 mangoes and 9 apples. If 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.

# of ways to succeed:: 8*3*9
# of random triples:: 20C3

P(condition) = 8*3*9/20C3 = 0.1895

Cheers,
Stan H.

Question 994185: what is the probability of drawing an ace or a king from a standard deck of cards
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
The probability of anything is the number of ways the event can occur that you would consider a success divided by the total number of ways that the event can occur. So if a success is an Ace or a King, how many Aces are there plus how many Kings? That's your numerator. The denominator is the total number of cards. You should reduce your fraction.
John
My calculator said it, I believe it, that settles it
Question 994182: Suppose that you work for a newly restructured automotive company with nearly 100,000 employees. You are in charge of purchasing engines from an overseas supplier. Company policy is that you purchase several hundred engines each month to be placed into cars on the assembly line.
Your overseas supplier of engines guarantees that 10% of the new engines shipped to you will have minor oil leaks that will require a slight modification before assembly. To check out the most recent monthly shipment of the engines, you randomly select and test 200 of these engines.What is the average number of leaky engines you would expect to get?
Answer by solver91311(20879) (Show Source):
Question 994154: find the area under curve between z=0.46 and Z=2.21
Answer by stanbon(69061) (Show Source):
Question 994060: Ally inherits $65,000 and decides to invest part of it in an education account for her daughter and the rest in a 10year CD. If the amount she puts in the education account is $12,000 more than twice the amount she puts in the CD, how much money does Ally invest in each account? Write an equation to solve this problem, if you could show how you came to it, and include units in this answer.
Thank you!
Answer by solver91311(20879) (Show Source):
Question 994007: gestation period, what is the probability that a baby will weigh between 6 and 9 pounds at 32 to 35 weeks
Answer by solver91311(20879) (Show Source):
Question 994081: Every now and then even a good diamond cutter has a problem and the diamond breaks. for one cutter, the rate of breaks is .1%.
If this cutter works on 75 stones, what is the probability that he breaks 2 or more?
I got 0.997420452.
Answer by rothauserc(2272) (Show Source):
You can put this solution on YOUR website! we use the binomial probability formula
Probability(P) = (nCk) * p^k * q^(nk) where nCk is the combination of n items taken k at a time
********************************************************************
P(k > or = 2) = 1  ( P ( k = 0 ) + P ( k = 1 ) )
p = .001, q = .999, n = 75
P ( k = 0 ) = 0.927708673390002
P ( k = 1 ) = 0.0696477983025527
P(k > or = 2) = 1  (0.927708673390002 + 0.0696477983025527)
P(k > or = 2) = 0.002643528
Question 994073: The television show NBC Sunday Night Football broadcast a game between the Colts and Patriots received a share of 22, meaning that among the TV sets in use, 22% were tuned to the game (based on Nielson data). An advertiser wants to obtain a second opinion by conducting its own survey, and a pilot survey begins with 20 households having TV sets in use at the time of that same NBC Sunday Night Football broadcast.
(+3)
a. Find the probability that none of the households are tuned to NBC Sunday Night Football (show with the table in excel and also do using the multiplication rule)?
(+3)
b. at least one is (show using the table using the direct method and the complement method)?
(+3)
c. exactly one household?
(+3)
d. find P(X≤3)=?
I got 0.006948520, 0.046145272, 0.039196760, 0.328910820.
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! None of the households would be 0.78^20 and your answer is correct.
At least one is the above answer subtracted from 1 or 0.9930. I used the complement method. At least 1 means not 0.
Exactly 1 is 20(0.22)(0.78^19)=0.0392 as you got.
For 2: 190*(0.22^2)(0.78^18)=0.1050
For 3: 1140(0.22^3)(0.78^17)=0.1777
I get basically the answer you got.
Question 994074: Ten peas are generated from parents having the green/yellow pair of genes, so there is a .75 probability that an individual pea will have a green pod. Find the probability that among the 10 offspring peas, at least 9 have green pods. Is it unusual to get this result (hint: compare it with the cutoff probability we established for the Rare Event Rule)?
I got 0.244025230.
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Ten peas are generated from parents having the green/yellow pair of genes, so there is a .75 probability that an individual pea will have a green pod. Find the probability that among the 10 offspring peas, at least 9 have green pods. Is it unusual to get this result (hint: compare it with the cutoff probability we established for the Rare Event Rule)?
I got 0.244025230.

Binomial Problem with n = 10 and p(green) = 0.75

P(9<= x <=10 = 1P(0<= x <=8) = 1  binomcdf(10,0.75,8) = 0.2440

Cheers,
Stan H.

Question 994072: a basketball player attempts 20 shots from the field during a game. this player hits about 35% of these shots. what is the chance the player hits more than 11 shots?
P(X>11)=?
I got this for the answer = 0.019579355, is this correct?
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! a basketball player attempts 20 shots from the field during a game. this player hits about 35% of these shots. what is the chance the player hits more than 11 shots?
P(X>11)=?
I got this for the answer = 0.019579355, is this correct?

Yes, that is correct.

Cheers,
Stan H.
Question 994059: In a sample of 1500 new iPhones, 9 were found to be defective. At this rate, how many defective iPhones would be expected in a sample of 100,000 new iPhones? Write a proportion to solve this problem and show all work.
Answer by rfer(15570) (Show Source):
Question 994021:
5. In the game of blackjack as played in casinos, the dealer has the advantage as most of the players do not play very well. As a result, the probability that the average player wins a hand is about 40%. An average player decides to play 10 hands. (15 points)
f. What is the probability that he will win in all hands? (3 points)
g. What is the probability that he will win at least 5 hands? (3 points)
h. What is the probability that he will win at least 1 hand? (3 points)
i. Assuming that this player has already lost the first 6 hands, what is the probability that he will win all remaining (4) hands? (3 points)
j. What is the average number of hands that an average player should expect to win in a 10hand game? (3 points)
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! 0.4^10=0.0001
=====
Probability wins 0=0.6^10=0.006
Probability wins 1=10*(0.4)*(0.6)^9=0.0403
wins 2= 10C2(0.4)^2)(0.6)^8=0.1209
wins 3=10C3(0.4)^3(0.6)^7=0.2150
wins 4=10C4(0.4)^4(0.6)^6=0.2508
These sum to 0.6330 This is what we don't want. The probability of winning at least 5 is everything else or 10.6330=0.3670
==========
This is the probability of 1probability of winning no hands, and that is 0.994
=========
That would be 0.4^4, since this is considered independent=0.0256
=========
Expected value is 10*0.4 or 4 hands.
Question 993924: The scores on the entrance exam at a wellknown, exclusive law school are normally distributed with a mean score of 125 and a standard deviation equal to 46. At what value should the lowest passing score be set if the school wishes only 2.5 percent of those taking the test to pass?
Set lowest passing score to
Answer by Theo(5548) (Show Source):
You can put this solution on YOUR website! you want a critical zfactor such that less than 2.5% will pass.
look up in the zscore table for a zscore that has an area to the left of it of (1  .025) = .975
that zfactor will be your critical zfactor and the value will be 1.96.
the formula for zscore is z = (x  m) / s
z is the zscore
x is the raw score
m is the mean
s is the standard deviation
since you know z and m and s, you can solve for x.
formula becomes:
1.96 = (x  125) / 46
solve for x to get x = 215.16
that's the score they need to pass.
here's a picture of what that looks like.
the first picture is the zfactor.
the second picture is the raw score.
the area of the normal distribution curve that is covered is shown in the picture and in digits at the bottom.
Question 993831: 1 2, 3, 4, 5, 6 NUMBERS OF PERSONS
.41, .30, .14, .11, .03, .01 RENT STABILIZED
Q. What is the variance of the number of person living in each unit?
rent stabilized?
So the way i tried was
(1^2)(0.41)+(2^2)(0.30)+(3^2)(0.14)+(4^2)(.11)+(5^2)(.03)+(6^2)(.01) =5.74
variance = E(x^2)[E(x)]^2 = 5.74(2.08)^2 = 1.1436 as my answer but it shows incorrect.
Can you please help me identify where i am wrong or is my answer put in different? Please help with answer. THanks
Answer by MathTherapy(4047) (Show Source):
You can put this solution on YOUR website!
1 2, 3, 4, 5, 6 NUMBERS OF PERSONS
.41, .30, .14, .11, .03, .01 RENT STABILIZED
Q. What is the variance of the number of person living in each unit?
rent stabilized?
So the way i tried was
(1^2)(0.41)+(2^2)(0.30)+(3^2)(0.14)+(4^2)(.11)+(5^2)(.03)+(6^2)(.01) =5.74
variance = E(x^2)[E(x)]^2 = 5.74(2.08)^2 = 1.1436 as my answer but it shows incorrect.
Can you please help me identify where i am wrong or is my answer put in different? Please help with answer. THanks
I got variance as: , which is what you should have gotten, but you seem to have copied the wrong value. You see, , not 1.1436 as you posted. I hope this helps!!
Question 993933: one thousand cars were stopped and tested for faults. The failure rate of two parts is as follows: Light bulbs  1 in 5 cars. Brakes  1 in 10 cars. What is the probabilty that one of the cars checked had faulty light bulbs but did not have faulty brakes?
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! These are independent
probability of the first occurring is 0.20
probability of the second NOT occurring is 0.90
The probability is their product, or 0.18.
Question 993973: In a binomial distribution, n = 12 and π = .60.
a. Find the probability for x = 5? (Round your answer to 3 decimal places.)
Probability
b. Find the probability for x ≤ 5? (Round your answer to 3 decimal places.)
Probability
c. Find the probability for x ≥ 6? (Round your answer to 3 decimal places.)
Probability
Answer by solver91311(20879) (Show Source):
Question 993926: In a bottlefilling process, the amount of drink injected into 16 oz bottles is normally distributed with a mean of 16 oz and a standard deviation of .32 oz. Bottles containing less than 15.86 oz do not meet the bottler’s quality standard. What percentage of filled bottles do not meet the standard? (Round the z value to 2 decimal places. Round your answer to 2 decimal places.)
P(x < 15.86) %
Answer by Theo(5548) (Show Source):
You can put this solution on YOUR website! the z factor is equal to (x  m) / s
x is the raw score.
m is the mean.
s is the standard deviation.
the critical z factor is the zscore of the cutoff value.
if the bottle contains less than 15.86 ounces of drink, then it is rejected.
that makes the cutoff z factor equal to (15.86  16) / .32 = .44
look up a zscore of .44 in the zscore table and you will get .33
that means that 33% of the zscores in the normal distribution table are below a zscore of .44.
that means that 33% of the bottles would be rejected if the amount of drink in the bottles was normally distributed with a mean of 16 ounces and the cutoff was any bottle that had less than 15.86 ounces of drink in it.
you can see the results in the following zscore calculator outputs.
the first output is based on the raw score.
the second output is based on the zscore.
the different in percentage is due to rounding of the zscore.
Question 993868: I'm not sure if this is the right place to post this but I'm pretty stuck. I'm having a hard time understanding how to do this or what I even need to do. Any help would be much appreciated.
Let's say you want to poll a random sample of 150 students on campus to see if they prefer to take online classes. Of course, if you took an actual poll you would only get one number (your sample proportion, phat). But, imagine all the possible samples of 150 students that you could draw and the imagined histogram of all the sample proportions from those samples.
1. What shape would the histogram of all the possible sample proportions (phat's) have?
2. Where would the center of that histogram be? (This answer should be a description in words based on what we know about phat sampling distributions.)
Now, given the information that about 35% of students actually prefer to take classes online respond to the following:
3. Discuss the conditions necessary to use the normal model here and explain whether or not they are met.
4. If you were to use the normal model for this phat sampling distribution, what would be the parameters (mean and standard deviation) for your model?
Answer by rothauserc(2272) (Show Source):
You can put this solution on YOUR website! A short discussion of p^ relative to this problem. The p^ symbol is used to refer to a proportion of the sample of the entire group. In this problem p^ would refer to the proportion of the sample group that prefer to take online classes.
********************************************************************************
1) Histogram will approximate a bell (normal) probability curve.
2) The mean of the sampling distribution is equal to the mean of the population.
3) In practice, a sample size of 30 is large enough when the population distribution is roughly bellshaped. Our sample size is 150 so this meets the constraint and the histogram approximates a normal(bell) probability curve.
4) P = 0.35, Q = 0.65, n = 150, then
mean = 150 * .35 = 52.5
std.dev. = square root( (0.35*0.65) / 150 ) = 0.038944405
Question 993833: given that person is in 36 and over age group, what is the probability that person applied to more than one school?
the data shows 51 for 36 and over and all together 808 total from person from diff age groups that applied to more than one school.
I thought for probability you are suppose to just take the 51 from all age group who applied to more than one which is 808. =.0631? But that is not it. Has to be 4 decimals as answer.
How do i solve this? Thanks
Answer by Theo(5548) (Show Source):
You can put this solution on YOUR website! what i believe you need is:
x = the number of people who are in the 36 and over age group that applied to more than 1 school.
judging from what you wrote, i believe that number will be 51.
y = the total number of people who are in the 36 and over age group, NOT the total number of people regardless of what age group they are in.
your probability should be x/y.
for example:
51 people in the 36 and over age group applied to more than one school.
372 people are in the 36 and over age group regardless of how many schools they applied to.
the probability that a person who is in the 36 and over age group applied to more than one school would be 51 / 372.
try it and see if you get the right answer.
let me know how you do.
if you don't get the right answer after doing that, then send me the complete question, as written, and i'll take a look at it for you.
also let me know what the answer is supposed to be.
Question 993825: A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them (USA Today, November 11, 2009). Suppose 10 employees are selected randomly and will be interviewed about company loyalty.
What is the probability that at least 2 of the 10 employees will say their company is loyal to them (to 4 decimals)?
The way I tried it was because it is at least 2 so have to do all numbers to 10. so P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)= So for this i looked on table adding 1.1664+.0064+.8464+3.6864+8.5264 and so on.. Obviously I am doing this wrong because the answer I put in shows it is incorrect. Can you tell me the formula and how to plug it in by steps? Thanks
Answer by rothauserc(2272) (Show Source):
You can put this solution on YOUR website! this problem uses the binomial probability(P) formula
P(k successes in n trials) = (combination of n trials with k successes) * p^k * q^(nk)
for this problem n = 10, k = (2, 3, 4, 5, 6, 7, 8, 9, 10), p = 0.25, q = 0.75
*******************************************************************************
then P(at least 2 employees say they are loyal) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 0.7559747695922852 approx 0.7560
Question 993777: What are the odds against an event happening when the probability of an event happening are 75%?
Answer by jim_thompson5910(33401) (Show Source):
You can put this solution on YOUR website! 75% = 3/4
there are 3 cases when the event happens out of 4 total
There is 1 case where the event does not happen
Odds against are 1:3
Question 993639: a) How many automobile license plate can be made if each plate contains 2 different letters followed by 3 different digits?
b) Solve the problems if the first digit cannot be 0.
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! a) How many automobile license plate can be made if each plate contains 2 different letters followed by 3 different digits?
26*25*10*9*8

b) Solve the problems if the first digit cannot be 0.
26*25*9*9*8
Question 993559: Dario puts 44 marbles in a box in which 14 are red, 12 are blue, and 18 are yellow. If Dario picks one marble at random, what is the probability that he selects a red marble or a yellow marble?
Answer by Fombitz(25151) (Show Source):
Question 993644: Find the number of ways in which 5 large books, 4 mediumsize books and 3 small books can be placed on a shelf so that all books with the same size are together?
Answer by fractalier(2141) (Show Source):
You can put this solution on YOUR website! Since all the books of the same size must be together, you effectively have but three groups, not twelve books...
Thus three groups, call them A, B, and C, can only be arranged in six ways:
ABC, ACB, BAC, BCA, CAB, and CBA.
Question 993647: Two cards are randomly drawn from an ordinary deck of playing cards without replacement. What is the probability that the first card is a heart and the second is a diamond?
Answer by fractalier(2141) (Show Source):
You can put this solution on YOUR website! Since this is an "and" problem, we must multiply the individual probabilities.
The first probability is 13/52 or 1/4.
The second probability is 13/51.
Multiplying them gives
1/4 x 13/51 = 13/204
Question 993635: Students from a statistics class were asked to record their heights in inches.
The heights (as recorded) were:
73, 74, 63, 64, 80, 72, 64, 64, 68, 75, 65, 65, 71, 72, 74, 50, 65, 68, 69, 66, 67, 67, 66
Are there any data values that may be suspected outliers? (Hint: remember this was the special section from 3.3)
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Students from a statistics class were asked to record their heights in inches.
The heights (as recorded) were:
73, 74, 63, 64, 80, 72, 64, 64, 68, 75, 65, 65, 71, 72, 74, 50, 65, 68, 69, 66, 67, 67, 66
Are there any data values that may be suspected outliers?
Looks like 80 might be an outlier.

Find the mean and standard deviation of the data.
Determine the zscore of 80

Follow your textbook's rule for identifying outliers.
Cheers,
Stan H.

Question 993638: An examination is designed where the students are required to answer 20 questions from a group of 25 questions. How many ways can a student choose the 20 questions?
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! An examination is designed where the students are required to answer 20 questions from a group of 25 questions. How many ways can a student choose the 20 questions?

Ans:: 25C20 = 25C5 = 25!/(20!)*5! = 53130

Cheers,
Stan H.

Question 993542: Three cards are drawn from a deck without replacement. What is the probability that all are jacks?
Answer by Edwin McCravy(13211) (Show Source):
Question 993502: The Better Business Bureau settles 65% of complaints it receives involving new car dealers. Suppose a sample of 90 complaints involving new car dealers is selected. Find the probability that Better Business Bureau settles less than 62 of these complaints
Please please help me
Thank you so much
Found 2 solutions by stanbon, Boreal: Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! The Better Business Bureau settles 65% of complaints it receives involving new car dealers. Suppose a sample of 90 complaints involving new car dealers is selected. Find the probability that Better Business Bureau settles less than 62 of these complaints

Binomial Problem with n = 90 and p(settle) = 0.65

P(x < 62) = binomcdf(90,0.65,61) = 0.7224

Cheers,
Stan H.
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! Looking at the problem, the expected number of complaints solved for 90 would be 90*0.65=58.5, and that is less than 62. Therefore, the probability will be fairly large that they settle fewer than 62.
The probability is <62/90=0.6888 (repeat)
1 sample proportion
z=(0.68880.65)/sqrt{ (0.65)(0.35)/90}
denominator is 0.0503
z= 0.03888/0.05
We want the probability that z is < than that number, The probability will be more than a half, since the sample proportion is greater than the purported probability.
That is 0.7817
Question 993420: 3.) If you were driving down the interstate at 60 mph and you look down at your cell phone for 3 seconds to see who texted you, how many feet did you travel down the interstate without your eyes on the road? Equate this distance
to a distance with which we are all familiar. Your distances have to be approximately the same.
Answer by josmiceli(13716) (Show Source):
Question 993421: 4.) Suppose that you could save $10 every hour, night, day. How long will it take to save 1 million dollars, rounded to the nearest tenth of a year? (use 1 year is 365 days)
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! 4.) Suppose that you could save $10 every hour, night, day. How long will it take to save 1 million dollars, rounded to the nearest tenth of a year? (use 1 year is 365 days)

It would take 100,000 hours.
= 100000/24 days
= 100000/(24*365) years
Question 628506: What NCAA college basketball conferences have the higher probability of having a team play in college basketball's national championship game? Over the last 20 years, the Atlantic Coast Conference (ACC) ranks first by having a team in the championship game 10 times. The Southeastern Conference (SEC) ranks second by having a team in the championship game 8 times. However, these two conferences have both had teams in the championship game only one time, when Arkansas (SEC) beat Duke (ACC) 76–70 in 1994 (NCAA website, April 2009). Use these data to estimate the following probabilities.
a. What is the probability the ACC will have a team in the championship game ?
b. What is the probability the SEC will have team in the championship game ?
c. What is the probability the ACC and SEC will both have teams in the championship game?
d. What is the probability at least one team from these two conferences will be in the championship game? That is, what is the probability a team from the ACC or SEC will play in the championship game?
e. What is the probability that the championship game will not have a team from one of these two conferences?
Answer by add84(1) (Show Source):
Question 993363: Based on the data provided, what is the expected cost of repairs over two years if each repair costs $60?
Answer by Fombitz(25151) (Show Source):
Question 993058: Find the probability of tossing a head on a coin or drawing a red ace from a standard deck of cards
Answer by vleith(2950) (Show Source):
You can put this solution on YOUR website! A coin has heads and tails. Two possibilities. The probability of getting heads is one chance in 2. 1:2
There are 52 cards in a standard card deck. Each suit has 13 different values. Each suit has one ace. Two of the suits are red. So there are 2 red aces
52 possibilities containing 2 red aces 2:52 or 1:26 (one chance out of 26)
Question 992907: Suppose Z is a standard normal random variable.
If P(z < Z < z) = 0.5588, find z.
Find P( 2.28 < Z < 0.63).
Please help me out!
Thank you so much
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Suppose Z is a standard normal random variable.
If P(z < Z < z) = 0.5588, find z.
Since 1/2 the probability is to the right of the mean
find the zvalue with a lefttail = 0.5+(1/2)(0.5588) = 0.7795
Ans: z = invNorm(0.7795) = 0.77051

Find P(2.28 < Z < 0.63) = normalcdf(2.28,0.63) = 0.2530

Cheers,
Stan H.

Question 992574: If X (bar) = 125, σ = 24, and n = 36, construct a 99% confidence interval estimate for the population mean, μ.
Answer by Fombitz(25151) (Show Source):

Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250, 20251..20295, 20296..20340, 20341..20385, 20386..20430, 20431..20475, 20476..20520, 20521..20565, 20566..20610, 20611..20655, 20656..20700, 20701..20745, 20746..20790, 20791..20835, 20836..20880, 20881..20925, 20926..20970, 20971..21015, 21016..21060, 21061..21105, 21106..21150, 21151..21195, 21196..21240, 21241..21285, 21286..21330, 21331..21375, 21376..21420, 21421..21465, 21466..21510, 21511..21555, 21556..21600, 21601..21645, 21646..21690, 21691..21735, 21736..21780, 21781..21825, 21826..21870, 21871..21915, 21916..21960, 21961..22005, 22006..22050, 22051..22095, 22096..22140, 22141..22185, 22186..22230, 22231..22275, 22276..22320, 22321..22365, 22366..22410, 22411..22455, 22456..22500, 22501..22545, 22546..22590, 22591..22635, 22636..22680, 22681..22725, 22726..22770, 22771..22815, 22816..22860, 22861..22905, 22906..22950, 22951..22995, 22996..23040, 23041..23085, 23086..23130, 23131..23175, 23176..23220, 23221..23265, 23266..23310, 23311..23355, 23356..23400, 23401..23445, 23446..23490, 23491..23535, 23536..23580, 23581..23625, 23626..23670, 23671..23715, 23716..23760, 23761..23805, 23806..23850, 23851..23895, 23896..23940, 23941..23985, 23986..24030, 24031..24075, 24076..24120, 24121..24165, 24166..24210, 24211..24255, 24256..24300, 24301..24345, 24346..24390, 24391..24435, 24436..24480, 24481..24525, 24526..24570, 24571..24615, 24616..24660, 24661..24705, 24706..24750, 24751..24795, 24796..24840, 24841..24885, 24886..24930, 24931..24975, 24976..25020, 25021..25065, 25066..25110, 25111..25155, 25156..25200, 25201..25245, 25246..25290, 25291..25335, 25336..25380, 25381..25425, 25426..25470, 25471..25515, 25516..25560, 25561..25605, 25606..25650, 25651..25695, 25696..25740, 25741..25785, 25786..25830, 25831..25875, 25876..25920, 25921..25965, 25966..26010, 26011..26055, 26056..26100, 26101..26145, 26146..26190, 26191..26235, 26236..26280, 26281..26325, 26326..26370, 26371..26415, 26416..26460, 26461..26505, 26506..26550, 26551..26595, 26596..26640, 26641..26685, 26686..26730, 26731..26775, 26776..26820, 26821..26865, 26866..26910, 26911..26955, 26956..27000, 27001..27045, 27046..27090, 27091..27135, 27136..27180, 27181..27225, 27226..27270, 27271..27315, 27316..27360, 27361..27405, 27406..27450, 27451..27495, 27496..27540, 27541..27585, 27586..27630, 27631..27675, 27676..27720, 27721..27765, 27766..27810, 27811..27855, 27856..27900, 27901..27945, 27946..27990, 27991..28035, 28036..28080, 28081..28125, 28126..28170, 28171..28215, 28216..28260, 28261..28305, 28306..28350, 28351..28395, 28396..28440, 28441..28485, 28486..28530, 28531..28575, 28576..28620, 28621..28665, 28666..28710, 28711..28755, 28756..28800, 28801..28845, 28846..28890, 28891..28935, 28936..28980, 28981..29025, 29026..29070, 29071..29115, 29116..29160, 29161..29205, 29206..29250, 29251..29295, 29296..29340, 29341..29385, 29386..29430, 29431..29475, 29476..29520, 29521..29565, 29566..29610, 29611..29655, 29656..29700, 29701..29745, 29746..29790, 29791..29835, 29836..29880, 29881..29925
