Answer to Question #183158 in Statistics and Probability for syam

Question

A sample size 25 is picked up at random from a population which is normally distributed with a mean of 100 and variance of 36. Calculate:-

(a) Pr{ ͞X≤99}

(b) Pr{98≤ ͞X≤100}

Answer 1

$n = 25 \\ \mu=100 \\ \sigma^ = 36 \\ \sigma = 6$

(a)

$z = \frac{\bar{X}- \mu}{\sigma / \sqrt{n}} \\ P(\bar{X} ≤ 99) = P(Z≤ \frac{99 -100}{6/ \sqrt{25}} )\\ = P(Z≤ -0.8333) \\ = 0.2024$

(b)

$P(98≤\bar{X} ≤ 100) = P(\bar{X}≤100) -P(\bar{X}≤98) \\ = P(Z≤ \frac{100-100}{6/ \sqrt{25}}) -P(Z≤ \frac{98-100}{6/ \sqrt{25}} )\\ = P(Z≤ 0) -P(Z≤ -1.666) \\ = 0.5 -0.0478 \\ = 0.4522$