Question

Let X be normally distributed with mean μ = 102 and standard deviation σ = 34. [You may find it useful to reference the z table.]


a. Find P(X ≤ 100). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

b. Find P(95 ≤ X ≤ 110). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

c. Find x such that P(X ≤ x) = 0.360. (Round "z" value and final answer to 3 decimal places.)

d. Find x such that P(X > x) = 0.830. (Round "z" value and final answer to 3 decimal places.)

Answer 1

Answer:

Given,

Mean = 102 and

Standard deviation = 34

a) P(x<100)=P(z<(100-102)/34)

= P(z<-0.06) [since from the normal distribution table]

P(x<100) = 0.5239

b) P(95<x<110)=P((95-102)/34<z<(110-102)/34)

=P(-0.21<z<0.24)

= P(z<0.24)-(1-P(z<0.21)) [since from the normal table]

we get

= 0.5948-(1-0.5832)

P(95<x<110) = 0.1780

c) P(X<=x)=0.360

By substituting the mean and standard deviation values we get x value

thus x=-0.36*34+102

x =89.76

d) P(X>x)=0.83

but actually we need the value of Z for P(X<x) So it is -0.96

thus x=-0.96*34+102

x = 69.36