Question

3. If you are taking a random sample from a normally distributed 5-12, determine if the following outcomes are a likely or unlike a. A sample mean greater than 106 for a sample of n4. b. A sample mean greater than 106 for a sample of n -36 population with ?-100 and ly event. (4 pts)

Answer 1

Here we have given that

$\mu$ = population mean =100

$\sigma$= population standard deviation=12

(A)

Now we want to calculate the probability of sample mean is greater than 106

for n=sample size =4

i.e $P( \bar x > 106)$

for finding this probability 1^st we need to find the Zscore

Z score -

=$\frac{106 - 100 } { \frac{12}{\sqrt(4)}}$

=1.00

i.e we get

P(Z > 1.00 ) = 1- P( Z < 1.00)

= 1 - 0.8413 using z standard normal table

=0.1587

we get

the probability of sample mean is greater than 106 =0.1587

i.e $P( \bar x > 106)$ =0.1587

(B)

Now we want to calculate the probability of sample mean is greater than 106

for n=sample size =36

i.e $P( \bar x > 106)$

for finding this probability 1^st we need to find the Zscore

Z score -

=$\frac{106 - 100 } { \frac{12}{\sqrt(36)}}$

=3.00

i.e we get

P(Z > 3.00 ) = 1- P( Z < 3.00)

= 1 - 0.9987 using z standard normal table

=0.0013

we get

the probability of sample mean is greater than 106 for n=36 is 0.1587

i.e $P( \bar x > 106)$ =0.0.0013