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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) = z = 53 – 50 = Find P(z<0.75)

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Find P(x45) X ̴ N( 50, 4²) so =50 and =4 P(Z<-1.25) = z = 45 – 50 = Find P(z<-1.25)

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P(Y>b)= Y ̴ N( 20, 9) so =20 and =3 From tables Z=1.5 z = b – = = b – = b – 20 b = 24.5

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The random variable X ̴ N(,3²) From tables Z= P= = 20 – =20– = Given P(X>20)=0.20, find the value of P z = 20 – =

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The random variable X ̴ N( 50, ²) From tables Z=0.8 so on graph Z=-0.8 P= = 46 –50 = -4/-0.8 = 5 Given P(X<46)=0.2119, find the value of P z = 46 – 50 = 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 = P(z < -1.43) P(z <1.43) = P(z < = =

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

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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 = – 510 = = 10.7

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b) Hence find the proportion of boxes that weigh more than 525g. P(W > 525) z = 525 – 510 = P(z > 1.40) = 1 – =

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Normal distribution calculator

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