91Ó°ÊÓ

i. The average weight of the $16$bags of$2$ounce candy in the sample, $\stackrel{¯}{x}=2$

ii. The standard deviation of candy bag weights is $2$ounces, $\mathrm{σ}=0.12$

iii. The number of bags of candy that have been chosen is $16$,$n=16$.

$x$ is the weight of one candy bag.

The average weight of 16 bags of candy was calculated from a sample of 16 bags has mean weight is$\stackrel{¯}{X}$.

The problem is solved using the normal distribution. As we know, the sample size is more than 30 for the population's standard deviation and the distribution of $N\left(2,\begin{array}{c}0.12\\ \sqrt{16}\end{array}\right)$

i. The confidence interval should be stated.

The confidence interval's output,

by calculated through

$$\begin{gathered} n=15 \\ CI=(1.9589,2.0411) \end{gathered}$$

ii. Below is a graph,

$1.96$ $2.04$

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

$EBM=\frac{U_{pper}\text{limit-lower limit}}{2}$

$EBM=\frac{2.0411-1.9589}{2}$

$EBM=0.0411$

1. The confidence interval should be stated.

By using calculator we get

$$\begin{gathered} n=16 \\ CI=(1.9418,2.0582) \end{gathered}$$

2.The graph is as follows:

$1.94$ $2.06$

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

$EBM=\frac{U_{\text{pper limit-lower limit}}}{2}$

$EBM=\frac{2.0582-1.9418}{2}$

$EBM=0.0582$

The region defined by the normal curve is used to compute the confidence interval. If all of the limits remain the same, the level of confidence remains the same. The level of confidence associated with the area then rises.

We are $98\%$ certain that the range $1.96to2.06$ comprises the mean weight for actual pollution candies.

As we increase the confidence level, we need to increase the sample size or the error bound, according to the error bound calculation, $206$people in the firm must be polled.