MATH 160 Name: ______

DIRECTIONS – PLEASE READ!!

  • Probability should be given to 4 decimal places.

Chapter 5:

1.Given the following problem, (a.) set up a probability distribution table & (b.) find the mean and standard deviation to two decimals.

The random variable x represents the number of foul shots a player makes when 2 shots are attempted. Assume that the player only misses an average of 31% of his/her free throws, and that the shots are independent.

Probabiliy that the shot is foul = 0.31
Probability that a shot is not foul = 1-0.31 = 0.69
X represents the number of foul shots when two shots are attempted. Then x is a random variable which assumes the values 0,1 and 2.
P(X = 0) = P(shots are not foul) = P(first shot is not foul)*P(second shot is not foul) = 0.69*0.69 = 0.4761
P(X = 1) = P(one shot is foul and one shot is not foul) = P(first is foul)*P(second is not foul) + P(first is not foul)*P(second is foul) = 0.31*0.69 + 0.69*0.31 = 0.2139 + 0.2139 = 0.4278
P(X = 2) = P(both shots are foul) = P(first is foul)*P(second is foul) = 0.31*0.31 = 0.0961
Probability table
x ..... P(X=x)
------
0 ..... 0.4761
1 ...... 0.4278
2 ...... 0.0961
------
Total ... 1.0000

2.) Exit polling has been a controversial practice in recent elections, because early release of the resulting information appears to affect whether or not those who have not yet voted will do so. Suppose that 83% of all registered CA voters favor banning the release of information from exit polls in presidential elections until after the polls in CA close. A random sample of 25 CA voters is to be selected.

a.) What is the probability that more than 19 voters favor the ban?

P(X > 19) = P(X=20)+...+P(X=25) = 0.7575

b.) What is the probability that at least 18 voters favor the ban?

P(X ≥ 18) = P(X=18) + ...+P(X=25) = 0.9505

c.) What is the mean value of the number of voters who favor the ban?

(X) = np = 25*0.83 = 20.75

d.) What is the standard deviation of the number of voters who favor the ban?

Standard deviation of X = √(np(1-p)) = √(25*0.83*0.17) = √3.5275 = 1.8782

3.) A question on a proficiency test is multiple choice with four possible answers, one of which is correct. Assuming that all responses are random guesses, find the probability that among 15 test subjects,

a.) exactly 9 answer the question correctly.P(X = 9) = C(15,9) * (1/4)^9 * (3/4)^6 = 0.0034

b.) at most 6 answer the question correctly.P(X ≤ 6) = P(X=0) + ... + P(X = 6) = 0.9434

c.) at least 10 answer the question correctly.P(X ≥ 10) = P(X = 10) +...+ P(X = 15) = 0.0008

Chapter 6

4.) If a normal distribution has mean 21 and standard deviation 6, then
A) / the median is 21because in normal distribution mean is same as median.
B) / the median is 15
C) / the median is 27
D) / the medians are 15 and 27
SHOW WORK!!

5.) Give the type of distribution pattern that occurs when the majority of the data values fall to the left of the mean.Positively Skewed

6.) In the figure below, what is the area under the curve between and ?

total area under the curve is 1. the left half has 0.5, the right half has 0.5 the shaded area is smaller than 0.5

7.) / The average gas mileage of a certain model car is 26 miles per gallon. If the gas mileages are normally distributed with a standard deviation of 1.3, find the probability that a car has a gas mileage of between 25.8 and 26.3 miles per gallon. Make sure you include your work!!
mu = 26 and sigma = 1.3 from your question
Get the z scores:
z(25.8) = (25.8-26)/1.3 = -0.153846
z(26.3) = (26.3-26)/1.3 = 0.230769
Now, using a z table
prob(-0.153846 < z < 0.230769)
= 0.1524
8.) / If the standard deviation of a population is 38 and we take a sample of size 25, then the standard error (the standard deviation of the sample mean) is what? Show how you got this. 38/squareroot of 25= 7.6
9.) / A biologist estimates that 76% of deer in the region carry a certain type of tick. For a sample of 300 deer selected at random, what is the chance that 246 or fewer deer have this tick? Make sure you include your work!!
Population.76
Sample=300

246/300-.76squareroot300/.76*.24= 2.43

10.) / The average age of doctors in a certain hospital is 44.0 years old with a standard deviation of 8.0 years. If 36 doctors are chosen at random for a committee, find the probability that the mean age of those doctors is less than 44.40 years. Assume that the variable is normally distributed. sample mean = 45.0
sample SD = 8 /√36 = 1.33
z value = 44.40 - 44.0 / 1.33 = answer is .300.40 to 2 DP 30.37
from the left side of the z- table, reqd. probability = 0.6554
ans : (D) : 0.6554
11.) / Mrs. Smith's reading class can read a mean of 175 words per minute with a standard deviation of 20 words per minute. The top 7% of the class is to receive a special award. What is the minimum number of words per minute a student would have to read in order to get the award? Show how you got this.
100-7=.9300
1.47 is z score
1.47*20+175=204.4

Chapter 7

12.) A sample of 25 different payroll departments found that the employees worked an average of 296.4 days a year with a standard deviation of 23.8 days. What is the 90% confidence interval for the average days worked, , of all payroll departments? Give an interpretation for this confidence interval.

13.) A study of peach trees showed that the average number of peaches per tree was 500. The standard deviation of the population is 200. A scientist wishes to find the 90% confidence interval for the mean number of peaches per tree. How many trees does she need to sample to be accurate within 16 peaches per tree? Make sure you show how you got your answer.

14.) A food snack manufacturer samples 14 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.5 and the sample standard deviation is 0.55, find the 95% confidence interval of the true mean. Give an interpretation for this confidence interval.

15.) A random sample of 170 voters found that 40% were going to vote for a certain candidate. Find the 99% limit for the population proportion of voters who will vote for that candidate. Give an interpretation for this confidence interval.

16.) A report states that 57% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to within 3% with 95% confidence? Show your work.

17.) Find the 95% confidence interval for the variance of the heights of maple trees if a sample of 32 trees has a standard deviation of 8 feet.