Given that X is normally distributed with mean 100 and stand

Given that X is normally distributed with mean 100 and standard deviation 9, compute the following for n = 16. (a) Mean (Round your answer to the nearest integer.) and variance of X (Round your answer to two decimal places (e.g. 98.76)) (b)P ( X ) 100) Round your answer to three decimal places (e.g. 98.765). (d)P (94

Solution

a) Mean = 100 (Given in the question)

Variance = s² / n where s = sample standard deviation.

= (9*9) / 16

= 5.0625

Std.Dev = 2.25 = (Variance)^0.5

b)

P (X < 95) = P (Z < 95-100 / 2.25)

= P (Z < -2.222)

= 0.013

c) For the third part,

we need P( X > 100) [ i.e area to the right of 100]

The conversion is simple :

P (X > 100) = 1 - P(X < 100)

= 1 - 0.50

= 0.50

d)

We need P ( 94 < X < 102) i.e area between 94 and 102.

Here is what you need to do :

Find the entire area to the left of 102 i .e [ P( X < 102 ) ] [ remember \'<\' sign denotes area on the left of that value]

Find the entire area to the left of 94 i.e P(X < 94)

The difference between the 2 will give you the area between 94 and 102.

So, P (94 < x < 102)

= P (X < 102) - P (X < 94)

= 0.8129 - 0.0038

= 0.809

Hope this helps.

I suggest you try to revise these kinds of problems by keeping a normal distribution diagram in front of you.

Ask if you have any more doubts in this question.