How many standard deviations above the mean is the cutoff fo

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.

Solution

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.

As the IQR = Q3 - Q1, and the corresponding z scores are 0.675 and -0.675, then in z score, the length of the IQR is 0.675 - (-0.675) = 1.35.

Thus, 1.5IQR = 1.5*1.35 = 2.025 standard deviation from the mean. [ANSWER, 2.025]

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b)

u + 2.025sigma = 65 + 2.025*3.5 = 72.0875 in [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    74.45      
u = mean =    65      
          
s = standard deviation =    3.5      
          
Thus,          
          
z = (x - u) / s =    2.7      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.7   ) =    0.003466974

Thus, we expect 0.003466974*25000 = 86.67435 = 87 such women. [ANSWER]