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2. Suppose X is a normally distributed random variable with a mean of 85 and a standard deviation of 5. Obtain the following
math   Statistics-And-Probability   
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Answer #1

Solution:

Given:

X - Normal (u = 85,0 = 5)

Part a) Find:

P( 80 < X < 86.5 ) =.............?

Find z score for x = 80 and for x = 86.5

z=\frac{x-\mu }{\sigma }

z=\frac{80-85 }{ 5} = \frac{ -5 }{ 5} = -1.00

z=\frac{86.5-85 }{ 5} = \frac{ 1.5 }{ 5} = 0.30

thus we get:

P(80 < X < 86.5) = P(-1.00 < Z < 0.30

P( 80 < X < 86.5 ) =P( Z< 0.30) - P( Z< -1.00 )

Look in z table for z = 0.3 and 0.00 as well as for  z = -1.0 and 0.00  and find corresponding area.

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 0.3 0.4 0.5 5000 5398 .5793 .6179 .6554 .6915 .5040 .5438 .5832 .6217 .

P( Z < 0.30 ) = 0.6179

and

.09 1.00 01 .02 -3.4 . 03 0003 10002 -33 0 05 2005,0005 -3.2 0 07 | 0007.0006 -3.1 0 10.0009,0009 -2.0 0 12 0012.0012 -29 0 1

P( Z < -1.00 ) = 0.1587

thus

P( 80 < X < 86.5 ) =P( Z< 0.30) - P( Z< -1.00 )

P( 80 < X < 86.5 ) =0.6179 - 0.1587

\mathbf{{\color{DarkOrange} P( 80 < X < 86.5 ) =0.4592 }}

Part b) Find:

P( X > 76) =...........?

Find z score for x = 76

z=\frac{x-\mu }{\sigma }

z=\frac{76-85 }{ 5} = \frac{ -9 }{ 5} = -1.80

thus we get:

P( X > 76) =P( Z > -1.80)

P( X > 76) =1 - P( Z < -1.80)

Look in z table for z = -1.8 and 0.00 and find corresponding area.

.00 .01 .02 103 .08 .09 003 -3.4 -3.3 10005 1-3.2 10007 -3.1 CD10 -3.0 10013 -2.9 20119 1-2.8 026 -2.7 100 85 1-26 1007 -2.5

P( Z < -1.80 ) = 0.0359

thus

P( X > 76) =1 - P( Z < -1.80)

P( X > 76) =1 - 0.0359

\mathbf{{\color{DarkGreen} P( X > 76) =0.9641}}

Part c) Find:

P( X < 87.8) =.............?

Find z score for x =87.8

z=\frac{x-\mu }{\sigma }

z=\frac{87.8-85 }{ 5} = \frac{ 2.8 }{ 5} = 0.56

thus we get:

P( X < 87.8) =P( Z< 0.56)

Look in z table for z = 0.5 and 0.06 and find corresponding area.

z .00 .01 .02 ,03 .04 .05 .06 .07 .08 .09 0.0 .5000 0.1 .5398 0.2 .5793 .6179 0.4 .6554 0.5 6945 .5040 .5438 .5832 .6217 .659

P( Z <0.56)= 0.7123

thus

P( X < 87.8) =P( Z< 0.56)

\mathbf{{\color{DarkBlue} P( X < 87.8) = 0.7123}}

Part d) Find b such that:

P( X < b) = 0.7704

Look in z  table for Area = 0.7704 or its closest area and find corresponding z value.

Area 0.7704 corresponding to 0.7 and 0.04, thus z = 0.74

Now use following formula to find x value:

x = \mu + z \times \sigma

x = 85 + 0.74 \times 5

x = 85 + 3.7

x = 88.7

that is: b = 88.7

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