Membership in Mensa requires a score in the top 2% on a standard intelligence te
ID: 3362924 • Letter: M
Question
Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.
1-Find the minimum Wechsler IQ score that satisfies the Mensa requirement.
2-If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131.
3-If 4 subjects take the Wechsler test and they have a mean of 132 but the individual scores are lost, can we conclude that all 4 of them are eligible for Mensa? Explain.
Explanation / Answer
1) for 2 percentil critical value of z score =2.0537
therefore corresponding minimum Wechsler IQ score that satisfies the Mensa requirement =mean +z*std deviation
=100+2.0537*15 =130.81
2) here std error of mean =std deviation/(n)1/2 =15/(4)1/2 =7.5
therefore probability that their mean score is at least 131 =P(X>131)=1-P(X<131) =1-P(Z<(131-100)/7.5)
=1-P(Z<4.1333)=1-0.99998 =0.00002 ~ 00000
3) No ; as some students has exceptionally high individual score while some have lower score then MENSA eligibility score ; while there score are averaging out to 132/