Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is $156,000. Assume the standard deviation is $42,000. Suppose you take a simple random sample of 49 graduates. Round all answers to four decimal places if necessary.

1. What is the distribution of XX? XX ~ N(,)
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
3. For a single randomly selected graduate, find the probability that her salary is between $150,400 and $159,100.
4. For a simple random sample of 49 graduates, find the probability that the average salary is between $150,400 and $159,100.
5. For part d), is the assumption of normal necessary? NoYes

Answer 1

Solutiona)

Given:

µ = 156,000.

σ = 42000

X ~ N ( µ = 156,000. , σ = 42000 )

Solution b)

µ_X̅ = µ = 156,000.
σ_X̅ = σ / √ (n) =42000/√49 =6000

~ N ( µ = 156,000. , σ/ √n =6000 )

Solution c)

X ~ N ( µ = 156,000. , σ = 4200 )

the probability that her salary is between $150,400 and $159,100.
P ( 150400<X < 159100)

P ( (150400-156,000 ) / 42000 )< ( X - µ ) / σ ) < ( 159100-156,000 ) / 42000 ))
P ( -0.13<Z < 0.07 )
=P(z<0.07)-P(z<-0.13)

=0.08245

#P ( 150400<X < 159100)=0.08245

Part d)

P ( 150400<< 159100)

P ( (150400-156,000 ) / (42000/√(49) )< ( - µ ) / ( σ / √ (n)) < ( 159100-156,000 ) / (42000/√(49)))
=P(-0.93<z<0.52)
=P(z<0.52)-P(z<-0.93)

=0.6973-0.1753

=0.52198

P ( 150400<< 159100)=0.52

Part e)

NO, For part d) assumptions of normality is not necessary because, sampling distribution of sample mean is normal, hence not need to specify the normality of X̅.