Question #212677

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.


1
Expert's answer
2021-07-02T11:55:07-0400

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<32406)P(<32)=P(z<\frac{32-40}{6})

=P(z<1.33)=P(z<-1.33)

=0.0912=0.0912

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

P(>27)=1P(z<27406)P(>27)=1-P(z<\frac{27-40}{6})

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

=10.0151=1-0.0151

=0.9849=0.9849

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

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

=P(z<51406)P(z<42406)=P(z<\frac{51-40}{6})-P(z<\frac{42-40}{6})

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

=0.96660.6306=0.9666-0.6306

=0.336=0.336

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

P(<x)=0.45P(<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,

X406=0.12566\frac{X-40}{6}=-0.12566 Solving for x,

x=6(0.12566)+40x=6(-0.12566)+40

=39.246=39.246

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

P(x>)=0.13P(x>)=0.13

P(<x)=10.13=0.87P(<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,

X406=1.126391\frac{X-40}{6}=1.126391 Solving for x,

x=6(1.126391)+40x=6(1.126391)+40

=46.758=46.758


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