Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 481 and a standard deviation of 108. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 537.1.
P(X > 537.1) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If 12 of the men are randomly selected, find the probability that their mean score is at least 537.1.
P(M > 537.1) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

We have given that,

Scores ~N(481,108)

If 1 of the men is randomly selected,We have to find probability that his score is atleast 537.1

i.e. P(X > 537.1)

We know that,

To find we need to find z score first.

Formula to find z score is

Where, X : test score

So z score becomes

   

Z= 0.519 ( rounded to 3 decimal)

Use below excel function to find

=NORM.DIST(X,mean,standard deviation,TRUE)

Here we are using z score so take x= z score = 0.519

mean =0 and standard deviation=1

(because Z score ~N(0,1))

So excel function becomes,

=NORM.DIST(0.519,0,1,TRUE)

So we get ,

Then enter this value in formula of P(X> 537.1)

P(X >537.1) = 1- 0.6981 = 0.3019

So probability that men score is at least 537.1 is 0.3019

Question 2:

If 12 men are randomly selected we have to find probability that their mean score is is at least 537.1

i.e. P(M > 537.1)

We know that,

Where,

So mean score =M~N(481,31.17691)

We know that,

We have given that,

Scores ~N(481,108)

If 1 of the men is randomly selected,We have to find probability that his score is atleast 537.1

i.e. P(X > 537.1)

We know that,

To find we need to find z score first.

Formula to find z score is

Where, X : test score

So z score becomes

   

Z= 0.519 ( rounded to 3 decimal)

Use below excel function to find

=NORM.DIST(X,mean,standard deviation,TRUE)

Here we are using z score so take x= z score = 0.519

mean =0 and standard deviation=1

(because Z score ~N(0,1))

So excel function becomes,

=NORM.DIST(0.519,0,1,TRUE)

So we get ,

Then enter this value in formula of P(X> 537.1)

P(X >537.1) = 1- 0.6981 = 0.3019

So probability that men score is at least 537.1 is 0.3019

Question 2:

If 12 men are randomly selected we have to find probability that their mean score is is at least 537.1

i.e. P(M > 537.1)

We know that,

Where,

So mean score =M~N(481,31.17691)

We know that,

We have given that,

Scores ~N(481,108)

If 1 of the men is randomly selected,We have to find probability that his score is atleast 537.1

i.e. P(X > 537.1)

We know that,

To find we need to find z score first.

Formula to find z score is

Where, X : test score

So z score becomes

   

Z= 0.519 ( rounded to 3 decimal)

Use below excel function to find

=NORM.DIST(X,mean,standard deviation,TRUE)

Here we are using z score so take x= z score = 0.519

mean =0 and standard deviation=1

(because Z score ~N(0,1))

So excel function becomes,

=NORM.DIST(0.519,0,1,TRUE)

So we get ,

Then enter this value in formula of P(X> 537.1)

P(X >537.1) = 1- 0.6981 = 0.3019

So probability that men score is at least 537.1 is 0.3019

Question 2:

If 12 men are randomly selected we have to find probability that their mean score is is at least 537.1

i.e. P(M > 537.1)

We know that,

Where,

So mean score =M~N(481,31.17691)

We know that,

To find we need to find z score first.

Z score =

So z score = 1.799408 = 1.799( rounded to 3 decimal)

Use below excel function to find

=NORM.DIST(X,mean,standard deviation,TRUE)

Here we are using z score so take x= z score = 1.799

mean =0 and standard deviation=1

So excel function becomes,

=NORM.DIST(1.799,0,1,TRUE)

So we get,

Then enter this value in formula of P(M> 537.1)

P( M > 537.1) = 1- 0.963991 = 0.036009 =0.0360 ( rounded to 4 decimal)

So probability that their mean score is at least 537.1 is 0.0360