# Questions on Algebra: Probability and statistics answered by real tutors!

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 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

Question 759440: According to a report by Scarborough Research, the average monthly household cellular phone bill is $73. Suppose local monthly household cell phone bills are normally distributed with a standard deviation of$11.35.
a. What is the probability that a randomly selected monthly cell phone bill is more than $100? b. What is the probability that a randomly selected monthly cell phone bill is between$60 and $83? c. What is the probability that a randomly selected monthly cell phone bill is between$80 and $90? d. What is the probability that a randomly selected monthly cell phone bill is no more than$55?

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According to a report by Scarborough Research, the average monthly household cellular phone bill is $73. Suppose local monthly household cell phone bills are normally distributed with a standard deviation of$11.35.
a. What is the probability that a randomly selected monthly cell phone bill is more than $100? Enter =1-normdist(100,73,11.35,TRUE) on EXCEL, the result is . b. What is the probability that a randomly selected monthly cell phone bill is between$60 and $83? Enter =normdist(83,73,11.35,TRUE)-normdist(60,73,11.35,TRUE) on EXCEL, the result is . c. What is the probability that a randomly selected monthly cell phone bill is between$80 and $90? Enter =normdist(90,73,11.35,TRUE)-normdist(80,73,11.35,TRUE) on EXCEL, the result is . d. What is the probability that a randomly selected monthly cell phone bill is no more than$55?
Enter =normdist(55,73,11.35,TRUE) on EXCEL, the result is

Question 759322: Suppose the life of a particular brand of calculator battery is normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If 16 batteries are randomly selected from the population, use Table 4 (Standard Normal probability table), software, or the TI-calculator to find the probability that the sample mean life will be between 70 and 80 hours. Use the Central Limit Theorem.
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Suppose the life of a particular brand of calculator battery is normally distributed with a mean of 75 hours and a standard deviation of 10 hours.
If 16 batteries are randomly selected from the population, use Table 4 (Standard Normal probability table), software, or the TI-calculator to find the probability that the sample mean life will be between 70 and 80 hours. Use the Central Limit Theorem.
-----
t(70) = (70-75)/[10/sqrt(16)] = -5/[10/4] = -20/10 = -2
---
t(80) = (80-75)/[10/4] = +2
----
P(70< x-bar <80) = P(-2,2,15) = tcdf(-2,2,15) = 0.9361
================================
Cheers,
Stan H.
================================

Question 759341: Assume that the population proportion of all business professionals who select an airline based on price is 0.25. A random sample of 250 business professionals is to be taken for further study.
A) What is the shape, mean (expected value), and standard deviation of the sampling distribution of the sample proportion for samples of size 250?
B) What is the probability that our sample of 250 results in a sample proportion that falls within 4 percent (+/- .04) of the population value?

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Assume that the population proportion of all business professionals who select an airline based on price is 0.25. A random sample of 250 business professionals is to be taken for further study.
A) What is the shape, mean (expected value), and standard deviation of the sampling distribution of the sample proportion for samples of size 250?
----
Ans: shape = normal ; mean = np = 250*0.25 = 62.5 ; std = sqrt[0.25*0.75/250] = 0.0274
=======================
B) What is the probability that our sample of 250 results in a sample proportion that falls within 4 percent (+/- .04) of the population value?
--
z(0.25-0.04) = z(0.21) = (0.21-0.25)/0.0274 = -1.46
z(0.25+0.04) = z(0.29) = (0.29-0.25)/0.0274 = 1.46
----
P(condtion met) = P(-1.46< z < 1.46) = normalcdf(-1.46,1.46) = 0.8557
======================
Cheers,
Stan H.

Question 759357: a student takes a 15 question multiple choice exam with two choices for each question and quesses on each question. assume the variable is binomial. round the intermediate and final answers to three decimal places. find the probability of guessing 11 out of 15 correctly.
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a student takes a 15 question multiple choice exam with two choices for each question and quesses on each question. assume the variable is binomial. round the intermediate and final answers to three decimal places. find the probability of guessing 11 out of 15 correctly.
--------------------
Binomial Problem with n = 15 and P(correct) = 1/2
------------------
P(x = 11) = 15C11(1/2)^15 = 0.0417
===================================
Cheers,
Stan H.
=======

Question 759359: Not sure how to set this problem up. Thank you for any and all input.
Find the z-score for the standard normal distribution where:
Area = 0.05 in the left tail

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Find the z-score for the standard normal distribution where:
Area = 0.05 in the left tail
------
Using a TI-84 I get: invNorm(0.05) = -1.645
==============================================
Cheers,
Stan H.
===========

Question 759369: Find the population standard deviation (σ not S) for the data set: {75,80,90,95,75,85,55,80,90,95}

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Find the population standard deviation (σ not S) for the data set: {75,80,90,95,75,85,55,80,90,95}
----
Ans: 11.45
===============
Cheers,
Stan H.
====================

Question 759385: A printing company's bookbinding machine has a probability of 0.75 for producing a perfect book. This machine used to bind three books together.
1) Find the probability that all three books are perfect books?

2) Find the probability that at least one of the books are perfect?

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A printing company's bookbinding machine has a probability of 0.75 for producing a perfect book. This machine used to bind three books together.
1) Find the probability that all three books are perfect books?
----
Binomial Prob. with n = 3 and P(perfect( = 0.75
P(x = 3) = 0.75^3 = 0.4219
--------------------------------
2) Find the probability that at least one of the books are perfect?
P(at least one perfect) = 1 - P(all are defective)
----
= 1 - 0.25^3
----
= 0.9844
============
Cheers,
Stan H.
============

An insurance company changes a 21-year-old male a premium of $250 for a one year$50,000 life insurance policy. A 21-year-old male has a 0.998 probability of living a year.
From the prospective of a a 21-year-old (or his estate), what are the values of the two different outcomes?
The value if he lives is ___ dollars.
The value if he dies is ____ dollars.
What is the expected value for a 21-year-old male who buys the insurance?
The expected value is ____ dollars.
What would be the cost of the insurance if the company just breaks even instead of making a profit?

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The value if he lives is __-250__ dollars.

This is the amount he loses in a year if he doesn't die and doesn't get the $50,000 payout. ------------------------------------------------------------------------------- The value if he dies is __50,000__ dollars. Note: obviously he doesn't benefit if he's dead (his family/estate does). This is the amount that gets paid out to the family/estate if he dies. ------------------------------------------------------------------------------- Note: P(living) = "probability of living" and V(dying) = "value of dying" (see above) What is the expected value for a 21-year-old male who buys the insurance? E[X] = P(living)*V(living) + P(dying)*V(dying) E[X] = 0.998*(-250) + (1-0.998)*(50,000) E[X] = -149.5 The expected value is __-149.50__ dollars. This is the amount, on average, the policy holder expects to lose each year. ------------------------------------------------------------------------------- What would be the cost of the insurance if the company just breaks even instead of making a profit? Let x = cost of insurance (ie the premiums paid per year) If the company breaks even, they will not lose money and they will not profit. So if the company breaks even, then they expect to get$0 on average, so E[X] = 0.

E[X] = P(living)*V(living) + P(dying)*V(dying)

0 = 0.998*(x) + (1-0.998)*(50,000)

0 = 0.998*(x) + (0.002)*(50,000)

0 = 0.998*(x) + 100

-100 = 0.998*x

0.998*x = -100

x = -100/0.998

x = -100.200400801603

x = -100.20

x is negative because this is from the viewpoint of the policyholder (he loses this much every year when making the premium payment)

So the cost of insurance is roughly $100.20 every year if the company just breaks even instead of making a profit Question 759131: A class consists of 30 woman and 40 men. If a student is randomly selected, what is the probability that the student will be male? Answer by John10(267) (Show Source): You can put this solution on YOUR website! A class consists of 30 woman and 40 men. If a student is randomly selected, what is the probability that the student will be male? --------------------------------- P( a student is a male) = (number of choices is men)/(number of all possible choices) = (40C1)/ (70C1) = 40/70 = 4/7 Hope it helps you. John Question 759138: A survey of students at a university show that 45% have a job. In a random sample of 500 students find the mean and standard deviation for the number of students who have a job. P.s. this is an intro to statistics question can u please answer it step by step Answer by John10(267) (Show Source): You can put this solution on YOUR website! A survey of students at a university show that 45% have a job. In a random sample of 500 students find the mean and standard deviation for the number of students who have a job. ------------------------ Hi, This is a binomial distribution problem. You have n = 500, p = 0.45, q = 1- p = 1 - 0.45 = 0.55 Mean = n * p St.deviation = SQRT(n * p * q) You can plug the numbers in. Hope it helps. John Question 759079: A machine that manufactures automobile parts produces defective parts 13% of the time. If 9 parts produced by this machine are randomly selected, what is the probability that fewer than 3 of the parts are defective? Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places. Answer by reviewermath(605) (Show Source): You can put this solution on YOUR website! Q: A machine that manufactures automobile parts produces defective parts 13% of the time. If 9 parts produced by this machine are randomly selected, what is the probability that fewer than 3 of the parts are defective? Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places. ---------------------------------------------------------------------------- A: Enter =BINOMDIST(2,9,0.13,TRUE) on EXCEL, the result is . Question 759056: their are 5 questions and 4 answers to each what is the probability of guessing 1 right Found 2 solutions by MathLover1, stanbon: Answer by MathLover1(6815) (Show Source): You can put this solution on YOUR website! There is right answer out of possible answers, so probability correct is or % or . Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! their are 5 questions and 4 answers to each what is the probability of guessing 1 right ---- Binomial Problem with n = 5 and P(correct) = 1/4 ------ P(x = 1) = 5C1(1/4)(3/4)^4 = 5(1/4)(81/256) = 405/1024 = 0.3955 ========================= Cheers, Stan H. ====================== Question 759027: A coin is tossed 5 times. What is the probability that it lands heads up on the third toss? Plz and thank you Math Whiz:) Answer by Alan3354(31538) (Show Source): You can put this solution on YOUR website! A coin is tossed 5 times. What is the probability that it lands heads up on the third toss? ---------------- 1/2, the same as every toss. Question 759055: with a graduation rate of 98%.... If 14 students are randomly selected , find the probability that at least 13 of them graduated. Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! with a graduation rate of 98%.... If 14 students are randomly selected , find the probability that at least 13 of them graduated. ------ Binomial Problem with n = 15 and p(grad) = 0.98 P(13<= x<=14) = 1 - binomcdf(14,0.98,12) = 0.9690 ------------------- Cheers, Stan H. =================================== Question 759009: a set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. find the value that is +1 standard deviation away from the mean. Answer by solver91311(17077) (Show Source): You can put this solution on YOUR website! So you are standing on a number line at the point 45 (that's the "mean" part). You know that your steps are 8.3 units long (that's the standard deviation part). You take one step (that's the 1 in the +1 part) to the right (that's the + in the +1 part). Where are you? John Egw to Beta kai to Sigma My calculator said it, I believe it, that settles it Question 758878: If a bird's life is normally distributed with a mean of 9 months and a standard deviation of 3, what percent of the population lives less than 11 months. Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! If a bird's life is normally distributed with a mean of 9 months and a standard deviation of 3, what percent of the population lives less than 11 months. ----- z(11) = (11-9)/3 = 2/3 --- P(x < 11) = P(z < 2/3) = 0.7475 =============== Cheers, Stan H. =============== Question 758904: On a test, if 130 questions are answered and 33% of them are correct, what is the number of correct answers? Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! On a test, if 130 questions are answered and 33% of them are correct, what is the number of correct answers? ------- Ans: (33/100)130 = 4290/100 = 42.9 correct Round up to 43 =================== Cheers, Stan H. =================== Question 758912: a class of 45 is required to provide one fourth sheet of paper each.how many sheets of paper will the whole class need? Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! a class of 45 is required to provide one fourth sheet of paper each.how many sheets of paper will the whole class need? ---- Ans: 45(1/4) = 11 1/4 sheets =============================== Cheers, Stan H. ============ Question 758937: find the probability of drawing 2 spades one after the other without replacement from a well shuffled pack of 52 cards. Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! find the probability of drawing 2 spades one after the other without replacement from a well shuffled pack of 52 cards. ------ # of ways to succeed: 13C2 = (13*12)/(1*2) = 78 ---------- # of possible outcomes: 52C2 = (52*51)/(1*2) = 1326 -------------- P(2 spades) = 78/1326 ======================== Cheers, Stan H. ================= Question 758924: Not really sure where to start on this problem. Any help you can give me would be greatly appreciated. Determine whether the distribution is a discrete probability distribution? x,P(x) 0,0.21 1,0.28 2,0.02 3,0.28 4,0.21 Answer by reviewermath(605) (Show Source): You can put this solution on YOUR website! Q: Not really sure where to start on this problem. Any help you can give me would be greatly appreciated. Determine whether the distribution is a discrete probability distribution? x,P(x) 0,0.21 1,0.28 2,0.02 3,0.28 4,0.21 ------------------------------------------------------------- A: , the distribution is a discrete probability distribution because = 0.21 + 0.28 + 0.02 + 0.28 + 0.21 = . Question 758796: suppose you will be taking exams in english, chemistry and statistics tomorrow. further, you know from past experience that you have a 50% chance of getting an A on each exam let x be a random variable representing the number of A's you earn. show the probability distribution of X. Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! suppose you will be taking exams in english, chemistry and statistics tomorrow. further, you know from past experience that you have a 50% chance of getting an A on each exam let x be a random variable representing the number of A's you earn. show the probability distribution of X. P(x=0) = 0.5^3 = 0.125 = 1/8 P(x=1) = 3(0.5)^3 = 3/0.125 = 3/8 P(x=2) = 3(0.5)^3 = 3/0.135 = 3/8 P(x=3) = 0.5^3 = 0.125 = 1/8 ----------------------------- Expected # of A's = np = 3(1/2) = 1.5 ========================================= Or if you want to work it this way: E(A) = 0*(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = [0+3+6+3]/8 = 12/8 = 1.5 =========================================== Cheers, Stan H. Question 758786: Find the following probabilities for standard normal random variable of z for problems a-d I spent an hour working with this problem and struggled to get the right answers. a) P(z>1.70) b) P(z<-1.38) c) P(0.21≤z≤2.52) d) P(-2.59 Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! Find the following probabilities for standard normal random variable of z for problems a-d I spent an hour working with this problem and struggled to get the right answers. ----- Are you using a z-table that you don't understand. These problems should not be that difficult. I used my TI-84 calculator to get the following answers: --------------------------- a) P(z>1.70) = normalcdf(1.70,100) = 0.0446 --------------------------- b) P(z<-1.38)= normalcdf(-100,-1.38) = 0.0838 --------------------------- c) P(0.21≤z≤2.52) = normalcdf(0.21,2.52) = 0.4110 --------------------------- d) P(-2.59 Something is missing "d". Cheers, Stan H. ============== Question 758842: Find the number of possible 5 card hands that contain At Least 1 King...taken from a standard 52 card deck? (using combinations) Find the number of possible 5 card hands that contain At Most 1 diamond....taken from a standard 52 card deck? (using combinations) Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! Find the number of possible 5 card hands that contain At Least 1 King...taken from a standard 52 card deck? (using combinations) ------ # of possible 5-card hands: 52C5 # of 5-card hands with no kings: 48C5 ----- Ans: 52C5-48C5 = 2,404,380 ============================================ Find the number of possible 5 card hands that contain At Most 1 diamond....taken from a standard 52 card deck? (using combinations) -- Number of at most 1 diamond = total number - Number with no diamonds ---- Ans: 52C5 - 39C5 = 2,023,203 ===================== Cheers, Stan H. =============== Question 758780: I calculated a probability that must be incorrect. ******* Here is the problem statement: Given: You have 100 colored marbles in a bag; 25-Red, 25-Blue, 25-Green & 25-Yellow. Problem: What is the probability of drawing 4 marbles (at 1 time) of the same color? ******* Here is my solution: There are 25-Red + 25-Blue + 25-Green + 25-Yellow = 100 total marbles. Draw 4 marbles at a time. The possible outcomes = 100C4 = 825 The number of ways to draw 4-Red marbles out of 25 is 25C4 = 50. The number of ways to draw 4-Blue marbles out of 25 is 25C4 = 50. The number of ways to draw 4-Green marbles out of 25 is 25C4 = 50. The number of ways to draw 4-Yellow marbles out of 25 is 25C4 = 50. The total ways you can get 4 of the same colored marbles is 50+50+50+50 = 200 P(ALL THE SAME COLOR) = 200 / 825 = 24% This can't be correct! Can you help me understand the error in my calculation? On the other hand, if it's correct, could you explain why it is correct? Obviously, I have no confidence in my answer. Thank you. Answer by Edwin McCravy(8999) (Show Source): You can put this solution on YOUR website! I calculated a probability that must be incorrect. ******* Here is the problem statement: Given: You have 100 colored marbles in a bag; 25-Red, 25-Blue, 25-Green & 25-Yellow. Problem: What is the probability of drawing 4 marbles (at 1 time) of the same color? ******* Here is my solution: There are 25-Red + 25-Blue + 25-Green + 25-Yellow = 100 total marbles. Draw 4 marbles at a time. The possible outcomes = 100C4 = 825 No! 100C4 = 3921225 The number of ways to draw 4-Red marbles out of 25 is 25C4 = 50. No! 25C4 = 12650 The number of ways to draw 4-Blue marbles out of 25 is 25C4 = 50. No! 25C4 = 12650 The number of ways to draw 4-Green marbles out of 25 is 25C4 = 50. No! 25C4 = 12650 The number of ways to draw 4-Yellow marbles out of 25 is 25C4 = 50. No! 25C4 = 12650 The total ways you can get 4 of the same colored marbl es is 50+50+50+50 = 200 No! 12650+12650+12650+12650 = 50600 P(ALL THE SAME COLOR) = 200 / 825 = 24%. No! 50600/3921225 = 0.0129041307 = or about 1.29% You are calculating combinations incorrectly. If you have a TI-83 or 84, To calculate 100C4, Type 100 Press MATH Press left arrow key once Press 3 Type 4 read 100 nCr 4 Press ENTER read 3921225 Edwin Question 758747: There are 25 students in a class, 15 male and 10 female. The teacher randomly selects students to go to the board to work out problems. What is the probability that 4 males and 4 females will be selected? Answer by John10(267) (Show Source): You can put this solution on YOUR website! Hi, You can use the formula Probability = (number of male choices * number of female choices)/ all choices = [ (15C4) * (10C4)]/ (25C8) You can take it from there. Hope it helps John Question 758600: A company operates four machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the shift. Machine Shift A B C D 1 41 20 12 16 2 31 11 9 14 3 15 17 16 10 A. The number of breakdowns is dependent on the shift, because the test value 11.649 is less than the critical value of 12.592. B. The claim that the number of breakdowns is independent of the shift cannot be rejected, because the test value 11.649 is less than the critical value of 12.592. C. The number of breakdowns is dependent on the shift, because the p-value is .07. D. The number of breakdowns is independent of the shift, because the test value 12.592 is greater than the critical value of 11.649. Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! A company operates four machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the shift. Machine Shift A B C D 1 41 20 12 16 2 31 11 9 14 3 15 17 16 10 ---- Ho: row and column factors are independent Ha: they are not independent ---------------- I ran a Chi-Sq test on the data and got the following: test stat: Chi-Sq = 11.649 p-value:: 0.07 ----- conclusion: Since the p-value is greater than 5%, fail to reject Ho. ---- Answer: B ------------- Cheers, Stan H. ============== A. The number of breakdowns is dependent on the shift, because the test value 11.649 is less than the critical value of 12.592. B. The claim that the number of breakdowns is independent of the shift cannot be rejected, because the test value 11.649 is less than the critical value of 12.592. C. The number of breakdowns is dependent on the shift, because the p-value is .07. D. The number of breakdowns is independent of the shift, because the test value 12.592 is greater than the critical value of 11.649. Question 758602: Multiple myeloma or blood plasma cancer is characterized by increased blood vessel formulation in the bone marrow that is a prognostic factor in survival. One treatment approach used for multiple myeloma is stem cell transplantation with the patient’s own stem cells. The following data represent the bone marrow microvessel density for a sample of 7 patients who had a complete response to a stem cell transplant as measured by blood and urine tests. Two measurements were taken: the first immediately prior to the stem cell transplant, and the second at the time of the complete response. Patient 1 2 3 4 5 6 7 Before 158 189 202 353 416 426 441 After 284 214 101 227 290 176 290 At the .01 level of significance, is there sufficient evidence to conclude that the mean bone marrow microvessel density is higher before the stem cell transplant than after the stem cell transplant? A. No B. Yes C. Cannot Determine Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! Two measurements were taken: the first immediately prior to the stem cell transplant, and the second at the time of the complete response. Patient 1 2 3 4 5 6 7 Before 158 189 202 353 416 426 44 After 284 214 101 227 290 176 290 Dif(A-B)-126 -25 101 126 126 250 151 ---- mean of Dif: 86.14 std of Dif:: 123.7 At the .01 level of significance, is there sufficient evidence to conclude that the mean bone marrow microvessel density is higher before the stem cell transplant than after the stem cell transplant? --- Ho: ud = 0 H1: ud > 0 (claim) ----- t(86.14) = (86.14/[123.7/sqrt(7)) = 1.8424 ------ p-value = P(t > 1.8424 when df = 6) = tcdf(1.8424,100,6) = 0.0575 ---- Conclusion: Since the p-value is greater than 1%, fail to reject Ho at the 1% level of significance. Ans:: No =============== Cheers, Stan H. =============== Question 758653: Using a tree diagram, show all possible 'boy-girl' outcomes of a family having 3 children. Identify the probability of the family having (a) exactly 3 boys (b) at least 2 girls Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! Using a tree diagram, show all possible 'boy-girl' outcomes of a family having 3 children. 1st column of the tree: boy::::::girl::: 2 elements in 1st column ----------------- 2nd column of the tree: write boy and girl to the right of each element in 1st column::: 4 elements in 2nd column ----------------- 3rd column of the tree: write boy and girl to gthe right of each element is 2nd column::::8 elements in 3rd column ----------------- Identify the probability of the family having (a) exactly 3 boys ::: boy,boy,boy occurs once on the tree ::: P(bbb) = 1/8 ------------------- (b) at least 2 girls patterns: ggg,ggb,gbg,bgg::: P(>= 2 girls) = 4/8 ============ Cheers, Stan H. ============ Question 758654: Given two independent events, A and B, how would you find: (a) P (A or B) (b) P (A and B) Answer by stanbon(57984) (Show Source): You can put this solution on YOUR website! Given two independent events, A and B, how would you find: (a) P (A or B) = P(A)+P(B)-P(A and B) = P(A)+P(B)-P(A)*P(B) --------------------- (b) P (A and B) = P(A)*P(B) --------------- cheers, Stan H. ----- Question 758191: A researcher polled 230 freshman at a university and found that 110 of them were enrolled in a history class. Estimate the probability that a randomly selected freshman is enrolled in a history class Answer by tommyt3rd(1120) (Show Source): Question 758621: Please help I've been working on this for awhile now. thank you in advance. Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ n=70, p=0.75 μ = σ = μ - 2σ = μ + 2σ = Answer by John10(267) (Show Source): You can put this solution on YOUR website! Hi, Since you have n = 70 and p = 0.75 and q = 1 - 0.75 = 0.25 Mean = μ = np σ = SQRT(n * p * q) After you find the mean and standard deviation, you can plug into the two expressions. Hope it helps you John Question 758578: In recent year, grade 10 new york state public school students taking a geometry mathematics assessment test had a mean raw score of 41 out of 88 total possible points with a standard deviation of 8. Assume that the scores are normally distributed. a) Find the probability that a student had a score between 30 and 40. b) What percent of the students scored less than 45? c) if 108 students took the test, how many had a test score of 49 and higher? Answer by reviewermath(605) (Show Source): You can put this solution on YOUR website! Question: In recent year, grade 10 new york state public school students taking a geometry mathematics assessment test had a mean raw score of 41 out of 88 total possible points with a standard deviation of 8. Assume that the scores are normally distributed. a) Find the probability that a student had a score between 30 and 40. b) What percent of the students scored less than 45? c) if 108 students took the test, how many had a test score of 49 and higher? -------------------------------------------------------------------------- Answer: a. Enter =normdist(40,41,8,TRUE) - normdist(30,41,8,TRUE) on EXCEL, the result is . b. Enter =normdist(45,41,8,TRUE) on EXCEL, the result is OR 69.15%. c. Enter =108*(1 - normdist(49,41,8,TRUE)) on EXCEL, the result is . Question 758566: The average high-school teacher annual salary is$43,000. Let teacher salary be normally distributed with a standard deviation of $18,000. a. What percent of high school teachers make between$40,000 and $50,000? Round to 3 decimals and keep '0' before the decimal point. b. What percent of high school teachers make more than$80,000? Round to 3 decimals and keep '0' before the decimal point.

You can put this solution on YOUR website!
Question:
The average high-school teacher annual salary is $43,000. Let teacher salary be normally distributed with a standard deviation of$18,000.
a. What percent of high school teachers make between $40,000 and$50,000? Round to 3 decimals and keep '0' before the decimal point.
b. What percent of high school teachers make more than \$80,000? Round to 3 decimals and keep '0' before the decimal point.
---------------------------------------------------------------------------
a. Enter
=ROUNDUP(normdist(50000,43000,18000,TRUE) - normdist(40000,43000,18000,TRUE),3)
on EXCEL, the result is
b. Enter
=ROUNDUP(1 - normdist(80000,43000,18000,TRUE),3)
on EXCEL, the result is

Question 758542: If mu is 68 and sigma is 4 what is the probability that x is greater then 72

Question 758548: Pop can weights are normally distributed with a mean of 12 ounces and a standard deviation of 0.7 ounces. What is the probability that a pop can contains at most 11.5 ounces? Enter your answer as a decimal rounded to the nearest thousandths place.

You can put this solution on YOUR website!
Q:
Pop can weights are normally distributed with a mean of 12 ounces and a standard deviation of 0.7 ounces. What is the probability that a pop can contains at most 11.5 ounces? Enter your answer as a decimal rounded to the nearest thousandths place.
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A:
Enter =ROUND(normdist(11.5,12,0.7,TRUE),3) on EXCEL, the result is .

Question 758549: What sample size is needed to estimate the mean white blood cell count (in cell micro-liter)?
For the population of adults in the United States? Assume that you want 99% confidence that the sample mean is within 0.2 of the population mean. The population standard deviation is 2.5. Show formula please,thank you.

You can put this solution on YOUR website!
Q:
What sample size is needed to estimate the mean white blood cell count (in cell micro-liter)?
For the population of adults in the United States? Assume that you want 99% confidence that the sample mean is within 0.2 of the population mean. The population standard deviation is 2.5. Show formula please,thank you.
-----------------------------------------------------------------------
A:
Enter =roundup((-normsinv((1-.99)/2)*2.5/0.2)^2,0) on EXCEL,the result is .

Question 758488: If a person can select 3 presents from 10 presents under a christmas tree, how many different combinations are there?
You can put this solution on YOUR website!
If a person can select 3 presents from 10 presents under a christmas tree, how many different combinations are there?
----------
The 1st choice is 1 of 10.
Then 1 of 9, 1 of 8
--> 10*9*8 = 720
------------------------
but, choosing A, B & C is the same as B, C & A, so
= 720/(3*2*1)
= 120 combinations

Question 758476: if you have Z number of balls in a box which consists of red, green, white and blue balls. then what is the fraction of red balls in the box. i just want the formula to solve this problem.
You can put this solution on YOUR website!
the number of balls in a box is equal to Z.
the box consists of red, green, white, and blue balls.
if you allow x to be the number of green balls in the box, then the fraction of red balls in the box is equal to x/Z.

Question 758437: the marks for test 1 for a particular subject are normally distributed with a mean of 52 marks and variance 26. a student was selected from class. what is the probability that this student obtained
(i)at least 60 marks in test1
(ii) it was told that 5% of the student obtained at least x marks in test 1.calculated the value x

You can put this solution on YOUR website!
the probabiity that a student picked at random from the class had a score greater than or equal to 60 marks is .2164.
see the following picture from the normal distribution calculator that i used.

the score that 5% of the students obtained or got greater than is equal to 64.389.

see the following picture from the normal distribution calculator that i used.

this calculator can be found at the following link:

http://davidmlane.com/normal.html

you need to calculate the standard deviation which is the square root of the variance.

Question 758449: can a probability become negative in any case???if yes,than when will the probability become negative????

You can put this solution on YOUR website!
the probbility can only be between 0 and 1.
if an event has a probability of 0, it exists 0% of the time.
if an event has a probability of 1, it exists 100% of the time.
it can never be negative.

Question 758327: The complete cycle of a fly’s life is 80 days. It spends 10 days as an egg, 40 days in larval form, and lives as a mature adult for 30 days. At a randomly chosen time, what is the probability that a given fly will NOT be in its adult form?

You can put this solution on YOUR website!
The complete cycle of a fly’s life is 80 days. It spends 10 days as an egg, 40 days in larval form, and lives as a mature adult for 30 days. At a randomly chosen time, what is the probability that a given fly will NOT be in its adult form?
------------------------------
Ans: (10 + 40)/80 5/8
============================
Cheers,
Stan H.
=============

Question 758340: The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 135 and a variance of 4. The material is considered defective if the breaking strength is less than 131 pounds. What is the probability that a single, randomly selected piece of material will be defective?
You can put this solution on YOUR website!
The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 135 and a variance of 4. The material is considered defective if the breaking strength is less than 131 pounds. What is the probability that a single, randomly selected piece of material will be defective?
---------
z(131) = (131-135)/4 = -4/4 = -1
-------------------
P(x < 131) = P(z < -1) = normalcdf(-100,-1) = 0.1587
================
Cheers,
Stan H.
======================

Question 758360: 6 coins are tossed simultaneously.what is the probability of getting at least 4 heads?

You can put this solution on YOUR website!
6 coins are tossed simultaneously.what is the probability of getting at least 4 heads?
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Binomial Problem with n = 6 and p(head) = 1/2
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P(4<= x <=6) = 1 - binomcdf(6,1/2,3) = 0.3438
==============================
Cheers,
Stan H.
==================

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