Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 48; p = .4

A) $\sigma = 6.66$
B) $\sigma = 3.39$
C) $\boldsymbol { \sigma } = 0.98$
D) $\boldsymbol { \sigma } = 7.51$

Question

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 48; p = .4

A) $\sigma = 6.66$
B) $\sigma = 3.39$
C) $\boldsymbol { \sigma } = 0.98$
D) $\boldsymbol { \sigma } = 7.51$

Quizplus · Accepted Answer

The standard deviation $\sigma$ for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$, where $n$ is the number of trials and $p$ is the probability of success. Substituting $n = 48$ and $p = 0.4$ gives $\sigma = \sqrt{48 \cdot 0.4 \cdot (1 - 0.4)} = \sqrt{48 \cdot 0.4 \cdot 0.6} = \sqrt{11.52} \approx 3.39$.

Use the Given Values of N and P to Find σ=6.66\sigma = 6.66σ=6.66

Use the Given Values of N and P to Find $\sigma = 6.66$