Example 1.11 The 90th parcentile of the standard ncrmal distributian is that val

Question

Example 1.11 The 90th parcentile of the standard ncrmal distributian is that valus on the harizontal axis such that the area under the z curve to theSelact of the valua is A standard normal curve table gives for fixad z the arca under the standard normal curve to the left of z, whereas here we have the area and want thu value of z. This is the "inverse' problem to AL s z) = ? so the table is used in an inverse it ins identify the 9gth z percentile. Here 0.9001 lies at the intarsection of the row mancad 2.3 and column mariced 0.03, so the ”th percantile is (approximately) z- fapproximataly)z Shaded area-9900 curve 9th percentile Finding the 99th percentile ey symmetry, the first percentile is as far below 0 as the 99th is above , so equals (1% lies below the first and also above th: 99th). Shaded area = .In -2.33= 1st percentile 2.33=outh percentile

Explanation / Answer

Ans:

99th percentile means that area under the z curve to the left of the value is 0.99

P(Z<=z)=0.99

z=2.33

Now,if we want cut off z values for top 1% and lowest 1% i.e. middle 98%

top 1% means 99th percentile

lowest 1% means 1st percentile

P(Z<=z)=0.01

z=-2.33

P(Z<=z)=0.99

z=2.33

So,z=+/-2.33

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