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**

Let X be normally distributed with mean μ =
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