91影视

Given in the question that, a large candy machine has$15$percent of orange candies

$SRS$$=25$candy count.

Assume that the sample size for SRS is $n$and that the sample distribution is $p$.

So, $n=25$and

$p=15\%$

$=0.15$

The mean of a sample proportion's sampling distribution $\hat{p}$and the population proportion $p$are the same, i.e., ${渭}_{\hat{p}}=p$.

In the given formula, replace $p$with $0.15$.

${渭}_{\hat{p}}=0.15$

The mean is ${渭}_{\hat{p}}=0.15$since the sampling proportion is an unbiased estimate for the population proportion.

$15$percent of orange candies in the machine.

SRS $=25$candy count.

Assume the sample distribution is $p$and the sample size is $n$.

$p=15\%$

$=0.15$

And, $n=25$

Determine the standard deviation of the sampling distribution:

$\hat{p}$is ${蟽}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Substitute $0.15$for $p$and $25$for $\mathrm{n}$:

${蟽}_{\hat{p}}=\sqrt{\frac{0.15(1-0.15)}{25}}$

$=\sqrt{\frac{0.15脳0.85}{25}}$

$=\sqrt{0.0051}$

$鈮�0.0714143$

$15$percent of orange candies in the machine.

SRS $25$candy count.

Assume that the sample distribution be $p$and sample size for SRS be $n$.

$p=15\%$

And, $n=25$

The product of sample size and the sampling proportion that is, $np$and $n(1-p)$both are less than at least $10$then the distribution of the samples is roughly Normal.

In the expression $np$, substitute $0.15$for $p$and $25$for $n$.

$25脳(0.15)=3.75$

In the expression $n(1-p)$, substitute $0.15$for $p$and $25$for $n$.

$25(1-0.15)=25脳0.85$

$=21.75$

Both $np$and $n(1-p)$are fewer than $10$, indicating that the Normal distribution requirement has not been satisfied.

As a result, $\hat{p}$'s sampling distribution is not close to Normal.

$15$percent of orange candies in the machine.

SRS $25$candy count.

Assume the sample distribution is $p$and the SRS sample size is$n$.

$p=15$

And,

$n=75$

Know that the standard deviation of the sampling distribution of $\hat{p}$is ${蟽}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$.

Simplify the preceding equation by substituting $0.15$for $p$and $75$for $n$.

${蟽}_{\hat{p}}=\sqrt{\frac{0.15(1-0.15)}{75}}$

$=\sqrt{\frac{0.45脳0.55}{75}}$

$=\sqrt{0.0033}$

$鈮�0.057445$

Therefore, the standard deviation is $0.057445$.