Question

The ages of personnel in the accounting department of a large company are normally distributed with a standard deviation of 7 years. There is a 0.02275 probability that the age of any randomly chosen person in the department is less than 22 and 0.15866 probabilty that the age of any randomly chosen person is greater than 43. What is the mean of this distribution? (the table below may help you answer this question) APPENDIX C NORMAL DISTRIBUTION PROBABILITY TABLE (To use this table, determine your z-score by taking the distance from the point you are trying to measure to the mean and dividing it by the standard deviation of your distribution. Then locate the number on the left-hand row to the tenth place, and the column that contains its hundreth decimal. The number in the box in which they intersect is the probability that the next data sample will be less than the point you are measuring (if that point is below the mean), or greater than the point you are measuring (if that point is above the mean).) Z-score 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.1 0.4 0.6 10.500000.49601 0.49202 10.48803 0.4840510.48006 0.47608 0.4721070.4681210.46414 10.4601770.456200.45224 0.44828 0.444330.44038 0.436440.432510.4285870.42465 10.420740.41683 0.41294 0.4090510.405170.401290.39743 0.3935810.38974 0.38591 10.3820970.378280.37448 0.3707070.3669370.3631710.3594210.355690.3519710.348271 10.344580.340900.33724 10.33360 10.3299710.23636|0.3227610.31918 0.315610.312071 10.308540.305030.301530.2980610.29460 0.29116 0.28774 0.28434 0.28096 0.27760 T0.27425 T0.27093 0.26763 0.26435 70.2610970.2378570.25463 70.251430.2482510.24510 10.241960.238850.23576 10.23270T0.2296570.226630.22363 0.2206570.217700.21476 10.211860.208970.20611 0.2032710.2004510.19766 0.1948910.19215 0.18943 0.18673 10.1840610.181410.17879 0.1761910.17361|0.17106 0.1685310.16602 0.16354 0.16109 10.1586610.156251 0.1538610.15151 0.149170.14686 0.144570.14231|0.1400710.13786 [0.135670.133500.13136 0.1292470.12714 0.125070.12302 0.121000.119000.11702 0.115070.113140.111230.10935 0.10749 0.10565 0.10383 0.10204 0.10027 0.09853 10.09680 10.095100.09342 10.091760.090120.0885110.08692 10.085340.083790.082261 0.0807610.079270.07780 10.07636 10.074930.07353 0.0721570.07078 0.06944 0.06811 10.06681 10.0655210.06426 10.06301 70.061780.0605710.05938 10.0582110.05705102055921 0.054800.053700.05262 0.0515570.0505010.049470.04846 0.04746 0.0464870.04551 0.04457 0.04363 0.04272 0.04182 0.040930.04006 0.0392010.03836 0.0375410.03673 0.03593 0.0351510.03438 0.03362 0.03288 0.03216 0.03144 0.03074 0.03005 0.02938 10.02872 10.028071 0.02743 0.0268010.0261910.0255910.025000.02442 10.0238510.02330 0.0227570.02222 0.02169 0.02118 10.0206810.02018 0.01970 0.019230.01876 0.01831 1.2

Answer 1

we have,

$\sigma$ = 7

x = u + z*$\sigma$

a) P(x < 22) = 0.02275

corresponding z score from table = -2

Thus, 22 = u - 2*7

u = 36

b) P(x > 43) = 0.15866

corresponding z score from table = +1

Thus, 43 = u + 1*7

u = 36

Thus, correct answer is c. 36