Question

In a study of environmental lead exposure and IQ, the data was collected form 182 children in Rochester, NY. There IQ scores at age of 6 approximately follow a normal distribution with mean of 84.9 and standard deviation of 14.4. Suppose one child had an IQ of 125. The researchers would like to know whether an IQ of 125 is an outlier or not. Calculate the upper fence for the IQ data, which is the upper limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the upper fence.

upper fence:

Then, calculate the probability of obtaining an IQ score value of 125 or greater. Keep a precision level of four decimal places for the probability

probability of IQ score greater than the upper fence:

Answer 1

Solution:

We are given IQ scores are normally distributed.

Mean = 84.9

SD = 14.4

n = 182

Q3 = Mean + Z*SD

Z for Q3 = 0.67449

(by using z-table, find inverse value for probability 0.75, we know that area under Q3 is 0.75)

Q3 = Mean + Z*SD

Q3 = 84.9 + 0.67449*14.4

Q3 = 94.61266

Q1 = Mean + Z*SD

Z for Q1 = -0.67449

(by using z-table, find inverse value for probability 0.25, we know that area under Q1 is 0.25)

Q1 = 84.9 - 0.67449*14.4

Q1 = 75.18734

IQR = Q3 – Q1 = 94.61266 - 75.18734 = 19.42532

IQR = 19.42532

Upper fence = Q3 + 1.5*IQR

Upper fence = 94.61266 + 1.5*19.42532

Upper fence = 123.7506

Upper fence = 123.75

An IQ of 125 is an outlier, because it is greater than 123.75.

Calculate the probability of obtaining an IQ score value of 125 or greater.

P(X>125) = 1 – P(X<125)

Z = (X – mean)/SD

Z = (125 - 84.9) / 14.4

Z = 2.784722

P(Z<2.784722) = P(X<125) = 0.997321

(by using z-table)

P(X>125) = 1 – P(X<125)

P(X>125) = 1 – 0.997321

P(X>125) = 0.002679

Required probability = 0.002679