Assume that IQ is normally distributed, with mean 100 and standard deviation 15.
ID: 3172056 • Letter: A
Question
Assume that IQ is normally distributed, with mean 100 and standard deviation 15.
(a) What is the probability that a randomly selected persons IQ is over 120?
(b) Find the values of Q1, Q2, and Q3 for IQ.
(c) Find the values of lower = Q1 1.5(Q3 Q1) and upper = Q3 + 1.5(Q3 Q1) for IQ. Recall that these were cutos for outliers.
(d) Find the probability of an outlier for IQ for a singleperson based on your values for (c).
(e) If we randomly selected 10 people, what is theprobability their average IQ is over 105?
Explanation / Answer
a. P(X>120)=P[Z>(120-100)/15] [Substitute given values in Z score formula, Z=(X-mu)/sigma, where, X is raw score, mu is population mean, and sigma is population standard deviation]
=P(Z>1.33)
=1-0.9082
=0.0918 (ans)
b. Find Z score corresponding to area 0.25, 0.50, and 0.75 representating Q1 (25%) Q2 (50%) and Q3 (75%) respectively.
The Z scores are -0.68, 0.00, and 0.68 respectively. Substitute the values in following Z score formula to obtain the raw scores.
-0.68=(X1-100)/15, X1=89.8, Q1=89.8
0.00=(X2-100)/15, X2=100, Q2=100
0.68=(X3-100)/15, X3=110.2, Q3=110.2
c. Lower cutoff= Q1-1.5(Q3-Q1)=89.8-1.5(110.2-89.8)=59.2
Upper cutoff=Q3+1.5(Q3-Q1)=110.2+1.5(110.2-89.8)=140.8
d. Any value beyond the cutoff is called outlier.
P(X>140.8)=P[Z>(140.8-100)/15]
=P(Z>2.72)
=1-0.9967 [Look into Z table to find area corresponding to Z score, the Z table gives area under standard normal curve to the left of Z, therefore, subtract the area from 1]
=0.0033
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.