Geometry test question #6

A rental car company claims the mean time to rent a car on their website is 60 seconds with a standard deviation of 22 seconds.

Gelneren [198K]

Answer:

Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds

Step-by-step explanation:

State the null hypothesis

H0: u = 60against the claim

Ha u≠ 60 (this is a two tailed test)

Sample size n= 36

Sample mean=X`= 55

Population mean = u= 60

The significance level α = 0.05

Standard deviation= Sd = 22 seconds

Z= X`- u / Sd /√n

Z= 55- 60 / 22/√6

z= - 5/22/6

Z= -1.3635

The value of z from the table is Z∝/2= ±1.96

The critical region is less than - 1.96 and greater than 1.96.

Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds

4 0

3 years ago

) e number of viewers ordering a particular pay-per-view program is normally distributed. Past history shows that 33.00% of the

lubasha [3.4K]

Answer:

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Past history shows that 33.00% of the time fewer than 20,000 people order the program

This means that X = 20000 has a pvalue of 0.33. So when X = 20000, Z = -0.44.

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{20000 - \mu}{\sigma}$