Question

The mean life of a television set is 138 months with a variance of 324. If a sample of 83 televisions is randomly selected, what is the probability that the sample mean would be less than 143.4 months

Answer 1

Answer: 0.9968

Explanation:

Let
`\overline{X}` be the sample mean life of television.

Given:
`\mu=138,\ \ \ \sigma^2=324\Rightarrow\ \sigma=√(324)=18,\ \ \ , n=83`

The probability that the sample mean would be less than 143.4 months will be :

`P(\overline{X}<143.4)=P(\frac{\overline{X}-\mu}{(\sigma)/(√(n))}<(143.4-138)/((18)/(√(83))))\\\\=P(Z<2.733)=0.9969\ \ \ [\text{By p-value table}]`

hence, the required probability = 0.9968