Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with ?=101 and ?=15.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 91 ? (Reminder: proportions are between 0 and 1 - don't put in percentages!)

(b) Enter the score which marks the lowest 25 percent of the distribution.

(c) Enter the score which marks the highest 5 percent of the distribution.

Answer 1

And Given that, test scores for children are nosnally distributed with se = 101 and o=15 @ now consider Proportion of children who have test scoles above al = P(x291) = 2 R(+291) - P(x-4 al 101) - Plze-0.67 from the standard aloknal Table, we get P(Z -0.67) - 0.7475 p(x291) 0.7475 Let x, be the score which makes the lowest 257 of the distribution. & P(xLxD = 254.20.25 Plzc21)=0.25 Nolnal Table, we get from the standard for Z -0.674 , Plze z) 2.0.25 x -l x,-101 0.7475 0.674

X 90-883 x = -0.674) (15) + 10 a 20.883 . X, E 90.883 let xo be the score which marke the highest . 54 of the Distributions. - P(x7X₂) = 5) = 0.05 8 (Z7 Zz) = 0.05 from the Standard Normal Table, we get for = 1:645 ; P(ZZZ₂) ~ 0.05 #2 = Xeli - Xq- 10) - z 1.645 = x, =(15 ( 1.645) + 10) = 125.673 - K2= 125-622 |