Answer to Question #92852 in Statistics and Probability for ramleq

Question

A Business Management monthly test was administered to a class consisting of 40 students. The mean score was 70 marks with a standard deviation of 6. Calculate the probability that a student obtained this score:

More than 82 marks
Between 64 and 72 marks
Less than 50 marks

Answer 1

Define mean score m = 70, standard deviation $\sigma=6$, score obtained by a student x.

Probability $P(a<x<b)=P(\frac{a-m}{\sigma}<z<\frac{b-m}{\sigma})$, where z has standard normal distribution.

$P(82<x<\infty)=P(\frac{82-70}{6}<z<\infty)=\\ P(2<z<\infty)=1-P(z\leq2)=1-0.9772=0.0228$

$P(64<x<72)=P(\frac{64-70}{6}<z<\frac{72-70}{6})=\\ P(-1<z<1/3)=P(z< 1/3)-P(z\leq-1)=\\ 0.6306-0.1587=0.4719$

$P(-\infty<x<50)=P(-\infty<z<\frac{50-70}{6})=\\ P(-\infty<z<-20/6)=0.0004$

Answer: P(x > 82) = 0.0228, P(64 < x <72) = 0.4719, P(x < 50) = 0.0004.