Question

If a score are normally with the mean of 30 and deviation of s, what perce a greater than 30? l'equal to s a greater than 37 dl between do and 34 e normally distribution in of 30 and the standard ors, what percent of the scores is

Answer 1

distributed we are géven that scores are normally with M = 30, o=5 we have to find what percent of score is (a) greater than 30 le. P[ X >30] standardising above, we get P[ X-W 30- 4 P[X> 30] = p[Z7 30- 30 ] =P[ z>o] = 1 - P[ Z Lo] P[ X >30] = 1 0.5 (using table) [P[ *>30] 20.

than 30. thes, 50% of score is greater (b) equal to 30 PLX = 30] standardising above, we get PCK=30] = PI X=I= 3004] = P(Z = 30-30] = P[2 = 0] since, z is a continuous roue so the probability at any discrete point is zero. therefore, P[X = 30 ] = 0 (c) greater than 37 loe. P[x>37] standardising above, we get P[X> 377 = p[ X=4> 37-]

P[X7373 = P [Z > 34–30] =P[2> 7/ = PLZ>1.4) = 1- P[Z (1.4) = 1-0.9192 (PLX 737] = 0.0808 [using z-table] of scores are thees, 8.08% percent greater than 37. (d) between 28 and 34 le P[284 XL 34] standardising above, we get P[ 284 X2 343 = plot me < 34-] =P[72-30 222 34-30 ] =P[-0.42Z < 0.8]

P[98<x<34] = P[Z < 0.8] – PIZL-0.4] =0.7881 -0.3446 JP [28<X234] = 0.44 35 of scores are therefore, 44.35% between 28 and 34