## 1

know | ||||||

n | 1000 | Z = 1.645 | ||||

s | 5.9 | |||||

m | 245 | |||||

Construct a 90% CI | ||||||

lower number | upper number | |||||

subract margin of error | add your margin of error |

A sample of 1000 items has a population standard deviation of 5.9 and a mean of 245. Construct a 90 percent confidence interval for Î¼.

## 2

Know | ||||||

n | 250 | LN | UN | |||

mean | 33.65 | |||||

Sta dev | 1.85 | |||||

z | 1.96 | |||||

At the end of 2016, 2017, and 2018, the average prices of a share of stock in a portfolio were $34.75, $34.65, and $31.25 respectively. To investigate the average share price at the end of 2020, a random sample of 250 stocks was drawn and their closing prices on the last trading day of 2020 were observed with a mean of 33.65 and a standard deviation of 1.85. Estimate the average price of a share of stock in the portfolio at the end of 2020 with a 95 percent confidence interval.

## 3

know | ||||

u | ||||

sta dev | ||||

n | ||||

z | ||||

LN | UN |

A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are Âµ = 3.25 and Ïƒ = 0.57. Suppose a random sample of 5000 male students is selected and the GPA for each student is calculated. Find the interval that contains 99 percent of the sample means for male students.

## 4

Know | ||||||||||||||||

mean | ||||||||||||||||

sta dev | ||||||||||||||||

n | ||||||||||||||||

P (X > 33) | 0.561 | ??? | ||||||||||||||

0.439 | ||||||||||||||||

mean=32.5 | 33 | |||||||||||||||

sta dev=3.25 | ||||||||||||||||

0.439 |

A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean of 32.5 lb and standard deviation of 3.25 lb, respectively, then based on a sample size of 500 boxes, what is the probability that the average weight of the boxes will exceed 33 lb?

## 5

mean of a binomial distribution = | n * p = u | ||||||

variance of a binomial distribution = n p ( 1 – p ) = |

If n = 52 and p = 0.66, then the standard deviation of the binomial distribution is

## 6

P ( X > 6 customers arriving within a minute) | ||||||||||||

0.8558 | ||||||||||||

u = 4.3 | 6 | |||||||||||

Consider a Poisson distribution with an average of 4.3 customers per minute at the local grocery store. If X = the number of arrivals per minute, find the probability of more than 6 customers arriving within a minute.

## 7

An important part of the customer service responsibilities of a cable company is the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.75 that troubles in a residential service can be repaired on the same day. For the first 6 troubles reported on a given day, what is the probability that all 6will be repaired on the same day?

## 8

z score | In the text | 5-4 | |||

knowns | |||||

mean | |||||

std dev | |||||

n | |||||

P ( X > 2.85) |

An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern state. The amount of juice obtained from each of these apples is approximately normally distributed with a mean of 2.75 ounces and a standard deviation of 0.11 ounce. What is the probability that a randomly selected apple will contain more than 2.85 ounces?

## 9

z score | |||||

Z = (X-u) / Sta Dev | Solve for x | ||||

Knowns | |||||

mean | 45 | ||||

Sta Dev | 7 | ||||

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 45 minutes and a standard deviation of 7 minutes. Suppose that in an effort to provide better service to the public, the director of the local office is permitted to provide discounts to those individuals whose waiting time exceeds a predetermined time. The director decides that 10 percent of the customers should receive this discount. What number of minutes do they need to wait to receive the discount?

## 10

z score for sampling distribution | ||||

? | ||||

P ( x bar > 88) | z = ( x bar – u ) / sta dev / sq root of n | |||

= (88-88.5) / 5.75 | ||||

-0.0869565217 | ||||

=standardize * SQRT(n) |

A random sample of size 350 is taken from a population with mean 88.5 and standard deviation 5.75. Find P(x bar > 88).