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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

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

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Normal distribution calculator

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2.2 Standard Normal Calculations, cont.. Because all Normal distributions are the same once we standardize, we can find percentages under Normal curves.

2.2 Standard Normal Calculations, cont.. Because all Normal distributions are the same once we standardize, we can find percentages under Normal curves.

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