Respuesta :
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%
"76.06%" would be the probability that the sample will contain between 165 as well as 195 blacks.
Given values,
- p = 12%
= 0.12
- n = 1500
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:
https://brainly.com/question/13796006
