The metropolitan bus company claims that the mean waiting time for a bus during rush hour is less than 5 minutes. a random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. at α = 0.01, test the bus company’s claim. assume the distribution is normally distributed. state the sample mean and the standard deviation

Decision is reject the as There is sufficient evidence to support the claim that the mean time is less then 5 minutes.

Step-by-step explanation:

Consider the provided information.

The mean waiting time for a bus during rush hour is less than 5 minutes.

Thus the null hypothesis is:

A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At α = 0.01,

Now use the table to find the z value, where α = 0.01,

The value of n is 20, sample mean is 3.7, population mean is 5 and standard deviation is 2.1.

Now, use the formula: Standardized test statistic for z-scores =.

Substitute the respective values in the above formula.

As, the test statistic is in the reject interval, so reject

There is sufficient evidence to support the claim that the mean time is less then 5 minutes.

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75