Question

score on last try: 0.3 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. > next question get a similar question you can retry this question below assume that a procedure yields a binomial distribution with n = 821 trials and the probability of success for one trial is p = 0.21. find the mean for this binomial distribution. (round answer to one decimal place.) μ = 172.4 find the standard deviation for this distribution. (round answer to two decimal places.) σ = 11.4 use the range rule of thumb to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. enter answer as an interval using square - brackets only with whole numbers. usual values =

Explanation:

Step1: Recall standard - deviation formula

The formula for the standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$, where $n = 821$ and $p=0.21$.

Step2: Calculate $1 - p$

$1 - p=1 - 0.21 = 0.79$.

Step3: Calculate $np(1 - p)$

$np(1 - p)=821\times0.21\times0.79=821\times0.1659 = 136.1039$.

Step4: Calculate standard - deviation

$\sigma=\sqrt{136.1039}\approx11.66$.

Step5: Calculate minimum usual value

The minimum usual value is $\mu - 2\sigma$. Since $\mu = 172.4$ and $\sigma\approx11.66$, then $\mu - 2\sigma=172.4-2\times11.66=172.4 - 23.32 = 149.08\approx149$.

Step6: Calculate maximum usual value

The maximum usual value is $\mu + 2\sigma$. So, $\mu + 2\sigma=172.4+2\times11.66=172.4 + 23.32 = 195.72\approx196$.

Answer:

$\sigma\approx11.66$
usual values = $[149,196]$