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30. Find the following percentiles for the standard normal distribution. Interpolate where appropriate. a. 91st b. 9th c. 75th d. 25th e. 6th 31. Determine za for the following values of α: Answer Answer C. α .663

can you help solve these 2 exercises. 31 and 32.
explain please.thanks

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30)

Let the mean of the standard normal distribution be u 0 Let the standard deviation of the standard normal distribution be σ 1 For the standard normal distribution, it can be observed that z is the desired percentile for N (μ, σ*) Find the 91st percentile of the standard normal distribution. For the lowest 91%, the equation becomes, Using the standard normal distribution area tables, the z-score corresponding to an area of 0.9100 is 1.34. So, the equation becomes P(z 1.34)-0.(2) Comparing (1) and (2), the 91st percentile is, P, -1.34This is illustrated in the figure Shaded area = 0.91 0 1.341 91st percentileFind the 9st percentile of the standard normal distribution. For the lowest 996, the equation becomes, P(z< P) 0.0900(3) Using the standard normal distribution area tables, the z-score corresponding to an area of 0.0900 is-1.34. So, the equation becomes, P(z <-1.34)-0.09(4) Comparing (3) and (4), the 9th percentile is, P 1.34 This is illustrated in the figure Shaded area 0.09 1.341 O 9th percentileFind the 75th percentile of the standard normal distribution. For the lowest 75% the equation becomes, P(z <%)=0.7500 (5) Here, an area of 0.7500 is halfway between the two probabilities that do appear in 0.7486 and 0.7517. The corresponding z-scores are 0.67 and 0.68. So, the z-score is, 0.67+0.680.675 In the usual notation, P(250.675)= 0.7500 (6) Comparing (5) and (6) the 75h percentlie is, P,,-0.675This is illustrated in the figure,Shaded area 0.75 00.675 75th percentileFind the 25m percentle of the standard normal distribution. For the lowest 25% the equation becomes, P( < P) 0.2500.(7) Here, an area of 0.2500 is halfway between the two probabilities that do appear in 0.2514 and 0.2483. The corresponding z-scores are-0.67 and-0.68. So, the z-score is, -0.67+(-0.68) -0.675 In the usual notation, P(25-0675) = 0.2500 (8) Comparing (7) and (8), the 25th percentile is Р.,--0.675This is illustrated in the figure, Shaded area 0.25 0.675 0 25th percentileFind the Gh percentile of the standard normal distribution. For the lowest 6%, the equation becomes, P(e< P)-0.0600 Using the standard normal distribution area tables, the z-score corresponding to an area of 0.0600 is-1.55. So, the equation becomes P(z <-1.55)-0.06 (10) Comparing (9) and (10), the 9th percentile is, Po =-1.55 This is illustrated in the figure Shaded area -0.06 1.550 1.55 0 6th percentile

31)

a) Z1-00055 Zo.9945 2.54 (ans) b) Z1-0.09 Zo 1.34 (ans) ) Z1-0.663-Z o337 -0.42 (ans)

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