Question

Assume that class grades follow a normal distribution of mean μ = 75 and the variance σ2 =144.

a) Find the probability that an individual's grade is greater than 81.

b) What should be the interquartile range?

c)Suppose you select at random (and independently) 10 students. What is the probability that only two of these students have a grade greater than 75?

d) If you draw a sample of size n = 10 from the population of grades described in a), what would be the probability that the sample variance, s2, is greater than 150?

Answer 1

Ans:

mean=75

standard deviation=sqrt(144)=12

a)

z=(81-75)/12

z=0.5

P(z>0.5)=0.3085

b)

For middle 50%, z values are +/-0.6745

Q1=75-0.6745*12=66.9

Q2=75+0.6745*12=83.1

IQR=Q3-Q1=16.2

c)mean=75,so

P(x>75)=0.5

Use binomial distribution with n=10,p=0.5

P(two have grade greater than 75)=¹⁰C₂*0.5^2*(1-0.5)^8=0.0439

d)

Chi square=(10-1)*150/144=9.375

df=10-1=9

P(chi square>9.375)=CHIDIST(9.375,9)=0.4034