How many standard deviations above the mean is the cutoff for outliers? (I\'m as

Question

How many standard deviations above the mean is the cutoff for outliers? (I'm asking for a z-score.) Use the Q3 + 1.5(IQR) rule. For your convenience, the z-score for the 75th percentile is approximately 0.675. Give your answer to 3 decimal places.  

Suppose the average height of TAMU women is normally distributed with mean 65 inches and standard deviation 3.5 inches. What height is tall enough to be an outlier for TAMU women? (I'm asking for the cutoff value above 65 inches.) Use 3 decimal places. Hint: use your answer from above in your calculation.

There are about 25,000 women at TAMU. About how many do we expect to have a height above 74.45 inches? Hint: Let X = height of a randomly chosen TAMU woman. Then P(X>74.45) is the proportion of women at TAMU who are taller than 74.45 inches. Round up to the nearest integer.

Explanation / Answer

Question 1:

The outliers are beyond Q3 + 1.5*(IQR)

We are already given here that Q3 = 0.675

Therefore, Q1 = - 0.675

Therefore IQR = Q3 - Q1 = 0.675 - (-0.675 ) = 1.35

Therefore Q3 + 1.5*(IQR) = 0.675 + 1.5*1.35 = 2.7

Therefore the z - score required for outlier detection is 2.7

Question 2:

Here we are given that:

Mean = 65 inches and standard deviation = 3.5 inches.

Outlier bound is computed as: the z score of 2.7 as:

Mean + 2.7*Standard deviation = 65 + 2.7*3.5 = 74.45

Therefore the 74.45 inches here is tall enough to be an outlier.

Question 3:

P(X > 74.45 ) = P(Z > 2.7)

Getting the above probability from the standard normal tables, we get:

P(Z > 2.7 ) = 0.0035

Therefore the number of women who expect to have a height above 74.45 inches would be computed as:

= 0.0035*25000

= 87.5

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.