Question

Research shows that the contamination index at a local beach is normally distributed with mean 160 and standard deviation 20. o é o What is the probability that the index level exceeds 190? What is the probability that the index level is between 140 and 180? Assume that the Public Health Department closes the beach when the contamination index lies above the 97.5 percentile. What is the maximum contamination index level at which the beach remains open? Find the first Quartile, Q1.

Answer 1

Given,

The contamination index at a local beach is normally distributed with mean 160 and standard deviation 20.

To find:

a) what is the probability that the index level exceed 190?

Solution:

Here, For se=190 zz Hill o = 190-160 - 20 = 30 20 za 1.5 p (exceed 190 )= P (H4>190)=P(271.5) af-p(231.5) = 1-0.9332 P = 0·0668 b) What is the probability that the index is between tho and 480 ? Solution : Here, For x=240, 2-10-460 20 z=-20 20 [ Z=-1 Also, for 4 = 180 2-480-460 20 23 20 20

Zot P (between 140 and 180)=P(640 x < 180) = P(-1<231) = P(Z31)-P(25-1) = 0.8413-0.1587 P = 0-6826 What is the maximum contamination index level at which the beach remain open? Value of 2 - 97.5 zo Hall = 1.96 = X-160 201.96 X = 199.21 level is 199.2 The maximum containation index d) find the girst quartile, Gl. Solution: First quartile : G1=l1-0-674516) =460-0-6745-(20) 5560-13.49 G, = 146:51