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

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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

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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.
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t(70) = (70-75)/[10/sqrt(16)] = -5/[10/4] = -20/10 = -2
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t(80) = (80-75)/[10/4] = +2
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P(70< x-bar <80) = P(-2,2,15) = tcdf(-2,2,15) = 0.9361
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Cheers,
Stan H.
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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?
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Ans: shape = normal ; mean = np = 250*0.25 = 62.5 ; std = sqrt[0.25*0.75/250] = 0.0274
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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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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
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P(condtion met) = P(-1.46< z < 1.46) = normalcdf(-1.46,1.46) = 0.8557
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Cheers,
Stan H.

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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.
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Binomial Problem with n = 15 and P(correct) = 1/2
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P(x = 11) = 15C11(1/2)^15 = 0.0417
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Cheers,
Stan H.
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Find the z-score for the standard normal distribution where:
Area = 0.05 in the left tail
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Using a TI-84 I get: invNorm(0.05) = -1.645
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Cheers,
Stan H.
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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}
Round your answer to the nearest 10th.
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Ans: 11.45
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Cheers,
Stan H.
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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?
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Binomial Prob. with n = 3 and P(perfect( = 0.75
P(x = 3) = 0.75^3 = 0.4219
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2) Find the probability that at least one of the books are perfect?
P(at least one perfect) = 1 - P(all are defective)
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= 1 - 0.25^3
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= 0.9844
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Cheers,
Stan H.
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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.
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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.
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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.
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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

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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?


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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

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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.
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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

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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.
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A:
Enter =BINOMDIST(2,9,0.13,TRUE) on EXCEL, the result is .

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