stats hw
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 μ.
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 µ = 3.25 and σ = 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).