Answer to Question #184623 in Statistics and Probability for jeykey

Question

A normally distributed population has mean 1,214 and standard deviation 122.

a. Find the probability that a single randomly selected element X of the population is between 1,100 and 1,300.

b. Find the mean and standard deviation of 𝑥̅for samples of size 25.

c. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300.

Answer 1

We have given that,

$\mu = 1214$

$\sigma = 122$

a.)$P(1100<X<1300)$

$= P(\dfrac{1100-1214}{122}<Z< \dfrac{1300-1214}{122})$

$= P(-0.93<Z<0.70)$

$= 0.70580-0.176$

$= 0.58$

b.) Now we have $n= 25$

Sample mean$= \mu_x = \mu = 1214$

Sample standard devaition $= \sigma_x = \dfrac{\sigma}{\sqrt n} = \dfrac{122}{5} = 24.4$

c.) $P(1100<X<1300)$ for $n=25$

$= P(\dfrac{1100-1214}{24.4}<Z< \dfrac{1300-1214}{24.4})$

$= P(-4.67<Z<3.52)$

$= 1$