Answer to Question #212677 in Statistics and Probability for Khan

Question

1. Given a normal distribution with μ = 40 and σ = 6, find

(a) the area below 32;

(b) the area above 27;

(c) the area between 42 and 51;

(d) the x value that has 45% of the area below it;

(e) the x value that has 13% of the area above it.

Answer 1

Area under a normal curve is equivalent to Probabilities.

μ = 40 and σ = 6

(a) Area below 32

Area below 32 is P(<32)

$P(<32)=P(z<\frac{32-40}{6})$

$=P(z<-1.33)$

$=0.0912$

(b) Area above 27 is P(>27)

$P(>27)=1-P(z<\frac{27-40}{6})$

$=1-P(z<-2.167)$

$=1-0.0151$

$=0.9849$

(c) Area between 42 and 51 is P(42<x<51)

$P(42<x<51)=P(<51)-P(<42)$

$=P(z<\frac{51-40}{6})-P(z<\frac{42-40}{6})$

$=P(z<1.833)-P(z<0.333)$

$=0.9666-0.6306$

$=0.336$

(d) The x value that has 45% of the area below it

$P(<x)=0.45$

The z-score corresponding to 0.45 on the left tail is -0.12566 (From z tables or =NORM.S.INV(0.45) excel function). Thus,

$\frac{X-40}{6}=-0.12566$ Solving for x,

$x=6(-0.12566)+40$

$=39.246$

(e) The x value that has 13% of the area above it

$P(x>)=0.13$

$P(<x)=1-0.13=0.87$

The z-score corresponding to 0.87 on the left tail is 1.126391 (From z tables or =NORM.S.INV(0.87) excel function). Thus,

$\frac{X-40}{6}=1.126391$ Solving for x,

$x=6(1.126391)+40$

$=46.758$