Question

based on a survey of 32 randomly selected employees (anonymously, of course) the company has determined that the average amount of time spent texting over a one-month period is 173 minutes with a standard deviation of 66 minutes. what is the probability that the average amount of time spent using text messages is more than 199 minutes

Answer 1

Answer: 0.0129

Explanation:

Given : Sample size : n=32

The average amount of time spent texting over a one-month period is :
`\mu=173\text{ minutes}`

Standard deviation :
`\sigma=66\text{ minutes}`

We assume that the time spent texting over a one-month period is normally distributed.

z-score :
`z=(x-\mu)/((\sigma)/(√(n)))`

For x= 199

`z=(199-173)/((66)/(√(32)))\approx2.23`

Now by using standard normal table, the probability that the average amount of time spent using text messages is more than 199 minutes will be :-

`P(x>199)=P(z>2.23)=1-P(z\leq2.23)\\\\=1- 0.9871262=0.0128738\approx0.0129`

Hence, the required probability = 0.0129