The Wechsler Intelligence Scale for Children is approximately normally distribut

Question

The Wechsler Intelligence Scale for Children is approximately normally distributed with a mean of 100 and a standard deviation of 15. What intelligence test score will place a child in the top 10 percent? Please show your work and reference the normal distribution table provided below!

Secend Decimal Place in 008.08000120 0160 0199 029 00279 9319 0359 .0438047890517005570596 636 00675 90714 073 2 00793.832 0871 00910 048 987 1025 0.1064 9.1103 114 .1217 125581293 1331 01368 1406 01443 0140 151 4 015501591162801664 0170001736 017720100.144 1879 0191950 1985 0.2019 02054 6208 02123 0215721908222 2291 0.2324 02357 62389 2422 03454 92488 92517 02549 5611 0242 02673 62704 02734 0264 02794 02823 282 0.2910 0.2939 0.2%7 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 3 01179 0.225 7 0 2580 80288 9 03159 0.3186 03212 0.3238 0.3264 0.12” 03315 03340 0.3365 0.3389 03413 3438 03461 3485 3508 03531 03554 03577 3599 36 1 03643 .3665 03686937088372903490377003790 3810 3830 03849 386903888 93907 3925344 03962 03980 3997 040 3 04032 14 04192 449 4669408204099 0411504131 0414704162 04177 0.4207 04222 04236 04251 04204n9 04292 0.4306 04319 5 94333 18 0 445 04145 04357 04370 04382 04394 044064418 64c9 e4441 4463944749448404495 04505 04515 94525 4535 4545 1.7 0.4554 04564 04573 04582 04591 04500 oaan .4616 .4625 eaca 4690456 9466404671 0.467804686 49346 0 476 L9 04713 0.47 19 04726 04732 04738 04744 04750 0.4756 0.4781 04767 94772947780478304788479304798040 0480 4812 04817 4826 0483094834048380434204840485048540485 18 04641 10482 0486 4164 04689487148750488 04881 94854 4887489 2.3 OAS93 04106 0408 04001 e0004 00006 04009 0W1 1 .4913 04916 2.4 0.4918 0.4920 04022 0.4925 04927 04929 0.4931 0.4932 0.4934 04936 2.5 0,4938 0.4940 ·4941 u943 0445 04946 0.4645 amie uni e4952 0.4955 04056 04957 04959 a4960 o 4961 04062 ·400 04964 466 94967 980499 64970 049794972493494 2,6 0.0053 04087 04987 LASS 0.4988 04989 0.4989 64989 04990 04990 0.4991 04,91 0.499 0.4992 0H92 0.4992 0.4992 04993 04993 4993 04994849404994049940.499 4954995499 0.4995 04995 0.4m6 04996 049% 0.406 0.4996 049% 04997 4997 04997049974997049970499 4997 4997499 04990 04993 4 099 0998 0.4999 0499999 499894999 0.4999 0 4999 04999 04999 94999 4999 04999 7 94999

Explanation / Answer

Mean is 100 and s is 15

For top 10% we need to look for a value that is at 90% from the bottom of the curve, thus, the z value from normal distribution table is 1.28

thus answer is mean+z*s=100+1.28*15=119.2

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