Question

A population of values has a normal distribution with μ=77.7μ=77.7 and σ=35.8σ=35.8. A random sample of size n=240n=240 is drawn.

1. Find the probability that a single randomly selected value is between 74.2 and 82.6. Round your answer to four decimal places.
   P(74.2<X<82.6)=P(74.2<X<82.6)=
   
2. Find the probability that a sample of size n=240n=240 is randomly selected with a mean between 74.2 and 82.6. Round your answer to four decimal places.
   P(74.2<M<82.6)=P(74.2<M<82.6)=  

Answer 1

Solution:

a)

P(74.2 < X < 82.6) = P((74.2 - 77.7)/ 35.8) < (x - ) /  < (82.6 - 77.7) / 35.8) )

= P(-0.10 < z < 0.14)

= P(z < 0.14) - P(z < -0.10)

= 0.5557 -  0.4602 Using standard normal table,  

Probability = 0.0955

b)

n = 240

m = 77.7

= / n = 35.8 / 240 = 2.3109

P( 74.2 < M < 82.6) = P((74.2 - 77.7) / 2.3109<( - m) / < (82.6 - 77.7) / 2.3109))

= P(-1.51 < Z < 2.12)

= P(Z < 2.12) - P(Z < -1.51) Using standard normal table,  

= 0.9830 - 0.0655

= 0.9175

Probability = 0.9175