Question

A company produces packets of soap
powder that are labeled "Giant Size 32 Ounces." The
actual weight of soap powder in a box has a Normal
distribution with a mean of 33 oz. and a standard
deviation of 0.7 oz. 95% of packets actually contain
more than x oz. of soap powder. What is x?
(a) 31.60
(b) 31.85
(c) 32.88
(d) 34.15
(e) 34.40

Answer 1

Answer:

(b) 31.85

Explanation:

First, we need to know the z-score for a one-sided confidence interval with 95% coverage. Looking at a z-score table, this value is 1.645. Then:

$x = (z \cdot \sigma) + \mu = (-1.645 \cdot 0.7) + 33 = 31.85$

Answer 2

z-scores illustrates how far a data element is, from the mean of the dataset.

The value of x is (e) 34.40

The given parameters are:

$\mu = 33$ --- mean

$\sigma = 0.7$ --- standard deviation

$\alpha = 95\%$ --- confidence level

First, we determine the z-score from the confidence interval $\alpha = 95\%$

From the z score table, the corresponding z score of $\alpha = 95\%$ is:

$z = 1.960$

The value of x, is calculated using:

$z = \frac{x - \mu}{\sigma}$

Substitute values for z, $\mu$ and $\sigma$

$1.960 = \frac{x - 33}{0.7}$

Multiply both sides by 0.7

$0.7 \times 1.960 = \frac{x - 33}{0.7} \times 0.7$

$1.372 = x - 33$

Collect like terms

$x = 33 + 1.372$

$x = 34.372$

Approximate

$x = 34.40$

Hence, the value of x is (e) 34.40

Read more about z-scores at:

brainly.com/question/13299273