Question

I just need someone to check that I got the correct answer. I chose B.

The average IQ score follows a Normal distribution, with a mean of μ = 100 and a standard deviation of σ = 15. What is the probability that the mean IQ score of 25 randomly selected people will be greater than 105?

0
0.0478
0.50
0.68
0.9522

Your input is highly appreciated. Thanks in advance.

Answer 1

You have to find the z-score for 105 then obtain the tail probability since it is less than half of the distribution from here: http://math-stat.net/unit%20normal%20table.pdf

Answer 2

Oh I calculated using a normal Pieulative probability..should I show my work so that you can check?

Answer 3

Marsulative*

Answer 4

I think the filter won't let me say Cu mulative

Answer 5

Yes it won't.

Answer 6

Should I show my work so that you can check?

Answer 7

Sure

Answer 8

I did
lower limit --> 105
upper limit --> 999, 999
sample mean --> 100
standard dev of sample --> σ/√n = 3
normalcdf(105, 999999, 100, 3)
=0.0477

Answer 9

Hmm, the way I usually do these is as follows
\[\sigma_{m} = \frac{ \sigma }{ \sqrt n} = \frac{ 15 }{ \sqrt 25 } = 3\]
\[z = \frac{ M - \mu }{ \sigma_{m} } = \frac{ 105-100 }{ 3 } = \frac{ 5 }{ 3 } = 1.66\]
Going along with the pdf I showed you, the tail probability in accordance with the z-score of 1.66 is 0.0485

Answer 10

Smart approach that works as well.
Thank you for taking the time to help me out!