One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About 12% of American adults identify themselves as black. Suppose we take an SRS of 1500 American adults and let X be the number of blacks in the sample. Use a Normal distribution to estimate the probability that the sample will contain between 165 and 195 blacks.

Solution :

Given :

p = 12% = 0.12

n = 15..

Mean is defined as the product of a sample size n and the probability p such that :

[tex]$\mu_X = np = 1500 \times 0.12$[/tex]

             = 180

Standard deviation may be defined as the square of a product of the sample size n, the probability p and the probability 1-p :

[tex]$\sigma_X = \sqrt{np(1-p)}$[/tex]

[tex]$\sigma_X = \sqrt{1500 \times 0.12 \times (1-0.12)}$[/tex]

     ≈  12.5857

The z-score is given by :

[tex]$z=\frac {x- \mu}{\sigma } $[/tex]

[tex]$z=\frac {165-180} {12.5857 }$[/tex]

 ≈  -1.19  

[tex]$z=\frac {x- \mu}{\sigma } $[/tex]

[tex]$z=\frac{195-180} {12.5857 }$[/tex]

 ≈  1.19  

Now determining the corresponding probability using table  :

P (165 ≤ X ≤ 195 ) = P (-1.19 ≤ Z ≤ 1.19 )

                            = 1-2 x P(Z < -1.19)

                            = 1-2 x 0.1170

                            = 1-0.2340

                            = 0.7660

                             = 76.60%

Given values,

           = 0.12

The mean will be

→ [tex]\mu x = np[/tex]

By putting the values,

       [tex]= 1500\times 0.12[/tex]

       [tex]= 180[/tex]

The standard deviation will be:

→ [tex]\sigma x = \sqrt{np(1-p)}[/tex]

       [tex]= \sqrt{1500\times 0.12\times (1-0.12)}[/tex]

       [tex]= 12.5857[/tex]

The z-score will be:

→ [tex]z = \frac{x- \mu}{\sigma}[/tex]

     [tex]= \frac{165-180}{12.5857}[/tex]

     [tex]= -1.19[/tex]

and,

→ [tex]z = \frac{195-180}{12.5857}[/tex]

     [tex]= 1.19[/tex]

hence,

The probability will be:

→ [tex]P(165 \leq X \leq 195) = P(-1.19 \leq Z \leq 1.19)[/tex]

                                [tex]= 1-2\times P(Z < -1.19)[/tex]

                                [tex]= 1-2\times 0.1170[/tex]

                                [tex]= 1-0.2340[/tex]

                                [tex]= 0.7660[/tex]

                                [tex]= 76.60[/tex] (%)

Thus the answer above is right.  

Learn more about probability here:

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