Question

Recent studies indicate that the typical 50-year-old woman spends $350 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $45 per year. We select a random sample of 40 women. The mean amount spent for those sampled is $335. What is the likelihood of finding a sample mean this large or larger from the specified population

Answer 1

The likelihood is 0.9826.

Explanation:

Here mean amount or μ = 350

Standard deviation = 45

Sample size n = 40

Let X be the sample mean where X>335

P(X>335)=
`P(z>(335-350)/(45/√(40) ) )`

`P(z>(-15)/(7.12))`

`P(z>-2.11)`

`P(z<2.11)`

=>
`0.5+P(0<z<2.11)`

=>
`0.5+0.4826=0.9826`

Hence, the likelihood is 0.9826.