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1. A set of 50 data values has a mean of 28 and a variance of 4. a. Find the standard score (z) for a data value = 31. z= 31− 28 3 = = 1.5 2 4 b. Find the probability of a data value > 31. P(x > 31) = P(z > 1.5) = 0.0668 2. Find the area under the standard normal curve: a. to the right of z = 2.37 0.0089 (This value comes from an online calculator.) b. to the left of z = 2.37 0.9911 (This value comes from an online calculator. No other work to show) Note that the answers to part a and part b add up to 1.00, as they should. 3. Assume that the population of heights of female college students is approximately normally distributed with mean m of 67 inches and standard deviation s of 3.95 inches. Show all work. (A) Find the proportion of female college students whose height is greater than 63 inches. z= 63− 67 = −1.01266 3.95 P(x > 63) = P(z > -1.01266) = 0.8444 (B) Find the proportion of female college students whose height is no more than 63 inches. P(x ≤ 63) = 1 – P(x > 63) = 1 – 0.8444 = 0.1556 4. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.45 inches and a standard deviation of 0.45 inches. Show all work. (A) What percentage of the grapefruits in this orchard have diameters less than 7.1 inches? z= 7.1− 6.45 = 1.4444 0.45 P(x < 7.1) = P(z < 1.4444) = 0.9257 Convert this proportion to a percentage by multiplying by 100%: 0.9257 * 100% = 92.57% (B) What percentage of the grapefruits in this orchard are larger than 6.8 inches? z= 6.8 − 6.45 = 0.7778 0.45 P(x > 6.8) = P(z > 0.7778) = 0.2183 Convert this proportion to a percentage by multiplying by 100%: 0.2183 * 100% = 21.83% 5. Find the normal approximation for the binomial probability that x = 4, where n = 13 and p = 0.3. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook. µ = np = (13) ( 0.3) = 3.9 σ = npq = (13) (0.3) (0.7) = 1.6523 z1 = ( 4.5 − 3.9) = 0.3613 z1 = (3.5 − 3.9) = −0.2421 1.6523 1.6523 P(x = 4) ≈ P(3.5 < x < 4.5) = P(-0.2421 < z < 0.3613) = P(z < 0.3613) – P(z < -0.2421) = 0.6411 - 0.4044 = 0.2367 The value from the table for P(x = 5) is 0.180 The value from the table for P(x = 4) is 0.234 6. A set of data is normally distributed with a mean of 100 and standard deviation of 15. · What would be the standard score for a score of 70? z= 70 −100 = −2 15 · What percentage of scores is between 100 and 70? P(70 < x < 100) = P(-2 < z < 0) = P(z < 0) – P(z < -2) = 0.5 – 0.0228 = 0.4772 Convert this to a percentage by multiplying by 100%: 0.4772 * 100% = 47.72% · What would be the percentile rank for a score of 70? The percentage of scores below 70 is: P(x < 70) = P(z < -2) = 0.0228 0.0288 * 100% = 2.88% ≈ 3rd percentile.