91影视

The sample of $68$snack pieces for the proportion of Bam-Bam pieces is $P\text{'},$ and the bag of snacks for the proportion of Bam-Bam snacks is$X$.

The problem is solved using the normal distribution. As we all know, the sample size for the population's standard deviation is greater than 30, and the distribution is$N\left(P,\sqrt{\frac{pq}{n}}\right)$.

In this we calculate p

By using formula

$p^{\text{'}}=\frac{5}{\text{total number of smack pieces}}$

by Computing,

$p^{\text{'}}=\frac{5}{68}$

$p^{\text{'}}=0.0735$.

i. The confidence interval should be stated.

The confidence interval's output,

$n=58$

$CI=(0.00853,0.13853)$

ii. The graph is as follows:

iii. The formula is used to compute the error bound.

$EBM=z_{\frac{伪}{2}}\sqrt{\frac{p^{\text{'}}{q}^{\text{'}}}{n}}$

$EBM=\frac{20.04}{2}\sqrt{\frac{0.0735脳0.9265}{68}}$

localid="1651847585804" $EBM=0.0554$

Do you believe that six packages of fruit snacks will provide enough data to provide accurate results, It's unlikely, because the samples were obtained using a basic random sample.