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Question: In the standard normal distribution, what is the value of z for the 73rd percentile?
A ) 0.73
B ) 1.1
C ) 0.23
D ) 0.613
Steps to derive
1 It is required to find that value of z, to the left of which lies 73% of the area of the normal distribution.
2 The area of the standard normal distribution to the left of the mean (z = 0) is 0.5.
3 Now, the problem is to find the value of k (to the right of the mean) so that P (0 < z < k) = 0.73 - 0.5 = 0.23.
4 Refer to the table giving the area under the standard normal curve from 0 to z.
5 The value of z = k corresponding to the area 0.23 from the table is 0.613.
6 So, P (z < 0.613) = 0.73. That is the value of z for the 73rd percentile is 0.613.
Hence the right answer is Option D
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