Question

1) If the mean on an test was 78 and the standard deviation was 12 find the following:

1. What score you would need to be above 50% of the class?
2. What score you would have received if 75% of the class did better than you did?
3. What your score was if your Z score was 1.4?

2) On your second test you received a score of 85 and the related Z score was .93. If the mean was 79, what is the standard deviation for the test?

3) On a fire fighter physical fitness test a candidate must be in the top 25% of the students to obtain a job. If the mean on the test is 80 and the standard deviation is 4.8, what score must the student get on the test to obtain a job?

Answer 1

1 a.Let X denote the scores. Assuming the distribution of scores to be normal, X~ Normal (mean=µ=78,standard deviation =Ơ=12)

Let x be the score that is needed to be above 50% of the class.

That is we need to find x such that P(X > x) =0.50

Hence, P(X >x)=0.50

=> 1-P(X>x)=1-0.5

=> P(X ≤ x )=0.5

=> P( (X-µ)/Ơ  ≤ (x -µ)/Ơ) =0.5

=>P(Z≤(x-78)/12)=0.5 [ Z=(X-µ)/Ơ ]

=> Ø( (x-78)/12) =Ø(0.5)......(1)

Now, Ø(0.5) =0 [Since normal distribution is symmetric, 50% of the values falls below and above the mean 0 of the standard normal distribution].....(2)

Hence, from (1) & (2)

(x-78)/12 =0

=> x-78 =0

=> x=78

Ans : 78

1 b.

X~ Normal (µ=78,Ơ=12)

We ned to find the score x for which 75% of the class did better

That is we need to find x such that P(X > x) =0.75

Hence, P(X >x)=0.75

=> 1-P(X>x)=1-0.75

=> P(X ≤ x )=0.25

=> P( (X-µ)/Ơ  ≤ (x -µ)/Ơ) =0.25

=>P(Z≤(x-78)/12)=0.25 [ Z=(X-µ)/Ơ ]

=> Ø( (x-78)/12) =Ø(0.25)......(1)

We need to find the z value for which the area = 0.25 that will fall below that value from the standard normal table.

From the normal table, for area =1-0.25=0.75, the z value is 0.674

Hence for area =0.25, the z value will be -0.674 [ Since its the left side of the normal curve]......(2)

Hence, from (1) & (2)

(x-78)/12 = -0.674

=> x-78 = -0.674*12

=> x-78= -8.088

=> x =78-8.088

=. x=69.912

Ans : 69.9 (approx)

1 c. If Z score =1.4

and Since Z = (X-µ)/Ơ

i.e X= µ + Ơ*Z

Now since µ=78 and Ơ=12

X =78+12*1.4=94.8

Ans : 94.8

2. Now X= µ + Ơ*Z....(1)

Given, X=85, Z=0.93,µ=79.

So, from (1) 85=79+Ơ*0.93

=> Ơ*0.93 =85-79

=>Ơ*0.93 =6

=>Ơ =6/0.93

=>Ơ =6.4516

Ans : Standard deviation Ơ =6.45 (approx)

3.

X~ Normal (µ=80,Ơ=4.8)

We ned to find the score x for the student to get a job i.e score for top 25% of the students

That is we need to find x such that P(X > x) =0.25

Hence, P(X >x)=0.25

=> 1-P(X>x)=1-0.25

=> P(X ≤ x )=0.75

=> P( (X-µ)/Ơ  ≤ (x -µ)/Ơ) =0.75

=>P(Z≤(x-80)/4.8)=0.75 [ Z=(X-µ)/Ơ ]

=> Ø( (x-80)/4.8) =Ø(0.75)......(1)

We need to find the z value for which the area = 0.75 that will fall below that value from the standard normal table.

From the normal table, for area =0.75, the z value is 0.674.......(2)

Hence, from (1) & (2)

(x-80)/4.8 = 0.674

=> x-80= 0.674*4.8

=> x-80= 3.2352

=> x =80+3.2352

=. x=83.2352

Ans : 83.2 (approx)