Question

The IQ scores of college students are normally distributed with the mean of 120 and standard deviation of 10. If a random sample of 25 students is taken, what is the probability that the mean of this sample will be between 120 and 125. (49.38%)

Answer 1

Answer:

$1596041915589_blob.png$= 120, $1596041915614_blob.png$ = 10, n= 25

We want to find P( 120 <   < 125 )

T

Where P( 120 <  
T
< 125 ) = P(
T
< 125 ) - P(
T
< 120 )

first find P( < 125 )

T

formula for z-score is

11 o/vn

$z = \frac{125 - 120 }{ 10 / \sqrt{25}}$

z = 2.5

P( < 125 ) = P(z < 2.5 )

T

using normal z table we get the

P(z < 2.5 ) =  0.9938

P( < 125 ) = 0.9938

T

now find P( < 120 )

T

formula for z-score is

11 o/vn

$z = \frac{120 - 120 }{ 10 / \sqrt{25}}$

z = 0

P( < 120 ) = P(z < 0 )

T

using normal z table we get the

P(z < 0 ) = 0.5000

P( < 120 ) = 0.5000

T

P( 120 <  
T
< 125 ) = P(
T
< 125 ) - P(
T
< 120 )

P( 120 <   < 125 ) =  0.9938 − 0.5

T

P( 120 <   < 125 ) = 0.4938

T

Probability = 0.4938

multiply by 100 to convert it to percentage.

0.4938 * 100 = 49.38%