Question

Barbara and James are in two separate classes taught by Mrs. Oshinsky. In Barbara’s class, the mean grade on a poetry test is a 74 and the standard deviation is 8. In James’s class, the mean grade is a 79 and the standard deviation is 3. Barbara and James both earn a 92 on their tests.

(A) Which student performed better in comparison to his or her class? You must show calculations to justify your response, or no credit will be given.

(B) In James’s class, what grade is comparable to Barbara’s?

(C) In Barbara’s class, what grade is comparable to James’s?

(D) What is the 60th percentile grade for each of the two classes?

Answer 1

• In Barbara’s class, the mean grade on a poetry test is a 74 and the standard deviation is 8.
  Barbara's score was 92.
  Hence, Z-Score of Barbara is:
  
  92 - 74 ZB = 2.25 8
  
  
  n James’s class, the mean grade is a 79 and the standard deviation is 3.
  James's score was 92.
  Hence, Z-Score of James is:
  
  92 - 79 ZJ 4.333333333 3
  
  
  Hence, the Z-Score of James is greater than Z-Score of Barbara. Hence, James is better performing than Barbara though their absolute scores are same.
  
  
• Let
  B
  be the grade of Barbara in James's class.
  Now,
  $\tiny \frac{X_B-79}{3}=2.25$
  
  XB = 79 + 3 * 2.25
  
  $\tiny \Rightarrow X_B=85.75$
  
  Hence, Barbara's grade in James's class is 85.75 .
  
  
• Let $\tiny X_J$ be the grade of James in Barbara's class.
  Now,
  $\tiny \frac{X_J-74}{8}=4.333333$
  $\tiny \Rightarrow X_J=74+8*4.333333$
  
  -> X XJ = 108.67
  
  
  Hence, Barbara's grade in James's class is 108.67 .
  
  
• Let X be the random variable denoting the scores of Barbara's class. It is given that:
  $\tiny X \sim N(74,8^2)$
  
  We are to find x such that,
  $\tiny P(X<x)=0.6$
  $\tiny \Rightarrow P(\frac{X-74}{8}<\frac{x-74}{8})=0.6$
  $\tiny \Rightarrow P(Z<\frac{x-74}{8})=0.6$
  $\tiny \Rightarrow \Phi(\frac{x-74}{8})=0.6$
  
  Now,
  
  Φ(0.25335) 0.6
  
  
  Hence,
  $\tiny \Rightarrow \Phi(\frac{x-74}{8})=\Phi(0.25335)$
  $\tiny \Rightarrow \frac{x-74}{8}=0.25335$
  $\tiny \Rightarrow x=74+8*0.25335$
  $\tiny \Rightarrow x=76.0268$
  
  Hence, the $\tiny 60^{th}$ percentile of Barbara's class is 76.03 .
  
  
  
  Let Y be the random variable denoting the scores of James's class. It is given that:
  $\tiny Y \sim N(79,3^2)$
  
  We are to find x such that,
  $\tiny P(Y<y)=0.6$
  $\tiny \Rightarrow P(\frac{Y-79}{3}<\frac{y-79}{3})=0.6$
  $\tiny \Rightarrow P(Z<\frac{y-79}{3})=0.6$
  $\tiny \Rightarrow \Phi(\frac{y-79}{3})=0.6$
  
  Now,
  
  Φ(0.25335) 0.6
  
  
  Hence,
  $\tiny \Rightarrow \Phi(\frac{y-79}{3})=\Phi(0.25335)$
  $\tiny \Rightarrow \frac{y-79}{3}=0.25335$
  $\tiny \Rightarrow y=79+3*0.25335$
  $\tiny \Rightarrow y=79.76005$
  
  Hence, the $\tiny 60^{th}$ percentile of James's class is 79.76 .
  
  

I hope this clarifies your doubt. If you're satisfied with the solution, hit the Like button. For further clarification, comment below. Thank You. :)