Answer to Question #154219 in Statistics and Probability for Dinush hehsna

Question

)A normally distributed population has mean 25.6 and standard deviation 3.3

(a) Find the probability that a single randomly selected element X of the 

population exceeds 30.

(b)Find the mean and standard deviation of X¯ for samples of size 9.

(C)Find the probability that the mean of a sample of size 99 drawn from this 

population exceeds 30.

Answer 1

a) mean=25.6, sd=3.3

P(X>30)

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

=$\frac{30-25.6}{3.3}$ =1.33

P(z>1.33)=0.09176 from the z tables.

=0.09176

b) mean of $\bar X$

$\mu=\bar X=25.6$

Standard deviation

$s=\frac{\sigma} {\sqrt n}$

=$\frac {3.3}{\sqrt 9}$ =1.1

c) P(X>30) given n=99

Z=$\frac{30-25.6}{\frac{3.3}{\sqrt{99}}}$ =13.27

P(z>13.27)$\approx$ 0 from the standard normal tables.

=0