Question

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 5929 with a mean life of 941 minutes. If the claim is true, in a sample of 109 batteries, what is the probability that the mean battery life would be greater than 948.8 minutes? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = $\mu$ = 941

variance = $\sigma$ ² =  5929

standard deviation = $\sigma$ =77

n = 109

$\mu$$\bar x$ = 941

$\sigma$_{$\bar x$ =} $\sigma$ / $\sqrt$ n = 77 $\sqrt$ 109 = 7.3751

P ($\bar x$ > 948.8 )

= 1 - P ($\bar x$ <948.8 )

= 1 - P ( $\bar x$ -  $\mu$$\bar x$/$\sigma$_{$\bar x$} ) < ( 948.8 - 941 / 7.3751)

= 1 - P ( z < 7.8 / 7.3751 )

= 1 - P ( z <1.06 )

Using z table

= 1 - 0.8554

= 0.1446

Probability = 0.1446