Question

A population of values has a normal distribution with μ = 247 and σ = 62.2. You intend to draw a random sample of size n = 16. (a) Find the probability that a single randomly selected value is greater than 295.2. (b) Find the probability that a sample of size n= 16 is randomly selected with a mean greater than 295.2. Give your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Answer:

a) $z = \frac{295.2-247}{62.2}=0.772$

And using the normal distribution table or excel we got:

$P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006$

b) $z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119$

And we can use the normal standard table or excel in order to find the probability and we got:

$P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907$

Step-by-step explanation:

For this case we know that the random variable of interest is normally distributed with the following parameters:

$X \sim N (\mu = 247, \sigma =62.2)$

Part a

We want to find this probability:

$P(X>295.2)$

And we can use the z score formula given by:

$z = \frac{X-\mu}{\sigma}$

Replacing we got:

$z = \frac{295.2-247}{62.2}=0.772$

And using the normal distribution table or excel we got:

$P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006$

Part b

We select a random sample of size n = 16 and we try to find this probability:

$P(\bar X >295.2)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119$

And we can use the normal standard table or excel in order to find the probability and we got:

$P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907$