Because P ( z < 0.44) = 0.67, 67% of all z values are less than .44, and .44 is
ID: 3150895 • Letter: B
Question
Because P(z < 0.44) = 0.67, 67% of all z values are less than .44, and .44 is the 67th percentile of the standard normal distribution. Determine the value of each of the percentiles for the standard normal distribution below. (Hint: If the cumulative area that you must look for does not appear in the z table, use the closest entry.)
(Round all answers to two decimal places.)
(a) The 91st percentile = (Hint: Look for area .9100.)
(b) The 70.2th percentile =
(c) The 50th percentile =
(d) The 9th percentile =
Explanation / Answer
(a)The nearest z value to the area 0.9100 is 1.34.
That is P(z<1.34)=0.9100
Thus the 91st percentile is 1.34.
(b) The nearest z value to the area 0.702 is 0.53.
That is P(z<0.53)=0.702
Thus the 70.2st percentile is 0.53.
(c) The nearest z value to the area 0.5000 is 0.00.
That is P(z<0.00)=0.5000
Thus the 50th percentile is 0.00.
(d) The nearest z value to the area 0.09 is -1.35.
That is P(z<-1.35)=0.09
Thus the 9th percentile is -1.35.