Question

A statistical analysis is made of the midterm and final scores in a large course, with the following results:pnN Average midterm score = 65, SD 10, Average final score = The scatter diagram is football shaped. 65, SD 12, r 0.6 About what percentage of the class final scores above 70? b. A student midterm was 75. Predict his final score Suppose the percentile rank of midterm score was 95%, predict his percentile rank on the final d.) Of those whose midterm score was 70, about what percentage of final scores over 80? MAan i C. Score

Answer 1

Let X shows the midterm score and Y shows the final score.

(a)

The z-score for Y = 70 is

Y-y 70 65 0.42 12 Sy

The percentage of the class final scores above 70 is

P(Y > 70) = P(z > 0.42) = 1 - P(z <= 0.42) = 1 - 0.6628 = 0.3372

(b)

Here we have

65, sr10, r 0.60, y 65, sy 12

Slope of the regression equation is

12 0.72 10 Sy br 0.6 ST

Y intercept of equation is

ay-ba65- 0.72.65 18.2

Equation of regression line is

y=18.2+0.72a

The predicted value for x = 75 is

18.2+0.72 75 74.45

(c)

First we need z-score that has 0.95 area to its left. z-score 1.645 has 0.95 area to its left. So z score corresponding to percentile rank of midterm is

1.645

So z-score for percentile rank of final score is

Zyrz 1.645 0.60-0.987

So percentile rank for final score is equal to area left to z-score 0.987. So required percentile rank is

P(z < 0.987) = 0.8382 = 83.82%

(d)

65, sr10, r 0.60, y 65, sy 12

Here conditional distribution of Y given X=70 follows normal distribution with mean

$\mu_{Y|x}=\mu_{y}+\rho\cdot \frac{\sigma_{Y}}{\sigma_{X}}(x-\mu_{X})=65+0.60\cdot \frac{12}{10}(70-65)=68.6$

and standard deviation:

$\sigma_{Y|x}=\sqrt{\sigma^{2}_{Y}(1-\rho^{2})}=\sqrt{144\cdot (1-0.36)}=9.6$

So z-score for Y = 80 | X=70 is

$z=\frac{80-68.6}{9.6}=1.1875$

So required probability is

P(Y > 80 | X = 70) = P(z > 1.1875) = 1 - P(z <= 1.1875) = 0.1175

Answer: 11.75%