Download presentation

Presentation is loading. Please wait.

Published byJulian Nelson Modified over 4 years ago

1
Normal Distribution 2 To be able to transform a normal distribution into Z and use tables To be able to use normal tables to find and To use the normal distribution to answer questions in context

2
Standard Normal Distribution z = x - X ̴ N(, ²) can be transformed into a Z score where Z ̴ N( 0, 1²)

3
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)

4
Find P(x45) 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

5
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

6
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

7
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

8
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 Stevens criteria

9
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) = 0.9236 P(z < -1.43 = 1-0.923 = 0.1895

10
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

11
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 chocolates b) 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

12
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

13
Normal distribution calculator

Similar presentations

OK

If E and F are two independent events with P(E)=.2 and P(F)=.4 find: 1) P(E|F) 2) P(F|E) 3) P(E∩F) 4) P(EUF) 1).22..4 3).083).52 WARM UP:

If E and F are two independent events with P(E)=.2 and P(F)=.4 find: 1) P(E|F) 2) P(F|E) 3) P(E∩F) 4) P(EUF) 1).22..4 3).083).52 WARM UP:

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google