Use the given values of n=2112, $p = \frac { 3 } { 4 }$ to find the minimum value that is not significantly low, $\mu - 2 \sigma$ and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.
A) Minimum: 1478.7; maximum: 1689.3
B) Minimum: 1544.2 ; maximum: 1623.8
C) Minimum: 1564.1; maximum: 1603.9
D) Minimum: 1623.8; maximum: 1544.2

Question

Use the given values of  n=2112, $p = \frac { 3 } { 4 }$  to find the minimum value that is not significantly low, $\mu - 2 \sigma$ and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.
A) Minimum: 1478.7; maximum:  1689.3 
B) Minimum:  1544.2 ; maximum:  1623.8 
C) Minimum: 1564.1; maximum:  1603.9 
D) Minimum: 1623.8; maximum:  1544.2

Quizplus · Accepted Answer

The mean, $\mu$, of a binomial distribution is given by $\mu = np$ and the standard deviation, $\sigma$, is given by $\sigma = \sqrt{np(1-p)}$. For $n = 2112$ and $p = \frac{3}{4}$, $\mu = 2112 \times \frac{3}{4} = 1584$ and $\sigma = \sqrt{2112 \times \frac{3}{4} \times \frac{1}{4}} = \sqrt{396}$ which is approximately 19.9. Therefore, $\mu - 2\sigma = 1584 - 2 \times 19.9 \approx 1544.2$ and $\mu + 2\sigma = 1584 + 2 \times 19.9 \approx 1623.8$.

Use the Given Values of N=2112 p=34p = \frac { 3 } { 4 }p=43​

Use the Given Values of N=2112 $p = \frac { 3 } { 4 }$