Suppose that the mean value of interpupillary distance (the distance between the

Question

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm.

(a) If the distribution of interpupillary distance is normal and a sample of n =25 adult males is to be selected, what is the probability that the sample average distance for these 25 will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.)

-Will be at least 67 mm?

(b) Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.)

- Will be at least 67 mm?

Explanation / Answer

Here mean is 65 and standard deviation is 5

a. We need to find P(64<x<66)

Converting x to z we get

P(64-65/(5/sqrt(25))<z<66-65/(5/sqrt(25)))

=P(-1<z<1)

=P(z<1)-P(z<=1)

=0.8413-0.1587

=0.6826

Now we need to find atleast 67

i.e P(x>=67)

=1-P(x<67)

Converting it into z we get

1-P(z<67-65/(5/sqrt(5)))

=1-P(z<2)

=1-0.9772

=0.0228

b. Now here as n>30, using central limit theorem we get sample distribution is normal

So find P(64<x<66)

Converting this into z we get

P(64-65/(5/sqrt(100))<z<66-65/(5/sqrt(100)))

=P(-2<z<2)

=P(z<2)-P(z<=-2)

=0.9772-0.0228

=0.9544

Now we need to find atleast 65

e P(x>=67)

=1-P(x<67)

Converting it into z we get

1-P(z<67-65/(5/sqrt(100)))

=1-P(z<4)

=1-1

=0

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