Question

Suppose a simple random sample of size nequals64 is obtained from a population with mu equals 88 and sigma equals 8. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P (x overbar greater than 89.7 )​? ​(c) What is Upper P (x overbar less than or equals 85.7 )​? ​(d) What is Upper P (87.35 less than x overbar less than 90.5 )​?

Answer 1

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 88$

$\sigma_{\bar X}= 8$

b) $z=\frac{89.7-88}{\frac{8}{\sqrt{64}}}= 1.7$

$P(Z>1.7) = 1-P(Z$

c) $z =\frac{85.7-88}{\frac{8}{\sqrt{64}}}= -2.3$

$P(Z$

d) $z =\frac{87.35-88}{\frac{8}{\sqrt{64}}}= -0.65$

$z =\frac{90.5-88}{\frac{8}{\sqrt{64}}}= 2.5$

$P(-0.65$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 88, \sigma = 8$

We select a sample size of n = 64

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sample mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 88$

$\sigma_{\bar X}= 8$

Part b

We want this probability:

$P(\bar X>89.7)$

We can use the z score formula given by:

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

And if we find the z score for 89.7 we got:

$z=\frac{89.7-88}{\frac{8}{\sqrt{64}}}= 1.7$

$P(Z>1.7) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

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

And if we find the z score for 85.7 we got:

$z =\frac{85.7-88}{\frac{8}{\sqrt{64}}}= -2.3$

$P(Z$

Part d

We want this probability:

$P(87.35$

We find the z scores:

$z =\frac{87.35-88}{\frac{8}{\sqrt{64}}}= -0.65$

$z =\frac{90.5-88}{\frac{8}{\sqrt{64}}}= 2.5$

$P(-0.65$