Assume college women have heights with the following distribution? (inches): ?N(

Question

Assume college women have heights with the following distribution? (inches): ?N(6060?, 2.62.6?). Complete parts? (a) through? (d) below. a. Find the height at the 75th percentile. The 75th percentile is nothing. ?(Round to one decimal place as? needed.) b. Find the height at the 25th percentile. The 25th percentile is nothing. ?(Round to one decimal place as? needed.) c. Find the interquartile range for heights. The interquartile range is nothing. ?(Round to one decimal place as? needed.) d. Is the interquartile range larger or smaller than the standard? deviation? The interquartile range is ? smaller larger than the standard deviation.

Explanation / Answer

a) for 75th percentile ; critical z =0.6745

henec corresponding height =mean +z*std deviaiton =61.7 (try 61.8 if above comes wrong due to rounding)

b)25 th percentile height =mean +z*std deviaiton=58.3(try 58.2 if above comes wrong due to rounding)

c)

interquartile range =Q3-Q1 =58.3-61.7 =3.6 (TRY 3.5 for rounding error)

D)The interquartile range is  larger than the standard deviation

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