Normal Distribution 2

To be able to transform a normal distribution into Z and use tablesTo be able to use normal tables to find and To use the normal distribution to answer questions in context

Standard Normal Distribution

• z = x -

X ̴ N(, ²) can be transformed into a Z score where Z ̴ N(0, 1²)

Find P(x<53)

X ̴ N(50, 4²) so =50 and =4

P(Z<0.75) = 0.7734

z = 53 – 50 = 0.75 4

Find P(z<0.75)

Find P(x≤45)

X ̴ N(50, 4²) so =50 and =4

P(Z<-1.25) = 0.1056

z = 45 – 50 = -1.25 4

Find P(z<-1.25)

1-0.8944

P(Y>b)=0.0668

Y ̴ N(20, 9) so =20 and =3

From tables Z=1.5

z = b – 20 3

1- 0.0668=0.9332

0.0668

1.5 = b – 20 3

4.5 = b – 20 b = 24.5

The random variable X ̴ N(,3²)

From tables Z=0.8416

P=0.20

0.8416 = 20 – 3

2.5248 =20– = 17.4752

Given P(X>20)=0.20, find the value of

P z = 20 – = 0.20 3

The random variable X ̴ N(50, ²)

From tables Z=0.8 so on graph Z=-0.8

P=0.2119

-0.8 = 46 –50

= -4/-0.8 = 5

Given P(X<46)=0.2119, find the value of

P z = 46 – 50 = 0.2119

P=0.7881

The heights of a large group of women are normally distributed with a mean of 165cm and a standard deviation of 3.5cm. A woman is selected at random from this group.

A) Find the probability that she is shorter than 160cm.

Steven is looking for a woman whose height is between 168cm and 174cm for a part in his next film.B) Find the proportion of women from this group who met Steven’s criteria

H=heights of women H ̴ N(165, 3.5²)

A) Find the probability that she is shorter than 160cm.

Find P(x<160)

z = 160 – 165 = 0.75 3.5

P(z < -1.43)

P(z <1.43) = 0.9236P(z < -1.43 = 1-0.923 = 0.1895

H=heights of women H ̴ N(165, 3.5²) B) Find P(168<x<174)

z = 168 – 165 = 0.86 3.5

P(z >0.86) = 0.8051

z = 174 – 165 = 2.55 3.5

P(z < 2.55) = 0.9946

P(0.86 <z < 2.55) = 0.9946-0.8051 = 0.1895

Boxes of chocolates are produced with a mean weight of 510g. Quality control checks show that 1% of boxes are rejected because their weight is less than 485g.A) Find the standard deviation of the weight of a box of chocolatesb) Hence find the proportion of boxes that weigh more than 525g.

W=weights of box of chocolates W ̴ N(510, ²) P(W<485) = 0.01

P z <485 – 510 = 0.01

485 – 510 = -2.3263

= 10.7

b) Hence find the proportion of boxes that weigh more than 525g.

P(W > 525)

z = 525 – 510 = 1.40 10.7

P(z > 1.40) = 1 – 0.9192 = 0.0808

Normal distribution calculator