Question

A company produces boxes of soap powder. the weight of soap powder in a box follows a normal distribution with a mean of 33 ounces and a standard deviation of 0.8 ounces. what weight should be advertised on the front of the boxes so that only 3% of the boxes are underweight?

(a) 35.4 oz
(b) 31.5 oz
(c) 30.6 oz
(d) 34.5 oz
(e) 32.6 oz

Answer 1

To ensure only 3% of soap powder boxes are underweight with a mean of 33 ounces and standard deviation of 0.8 ounces, the advertised weight should be 32.6 ounces. Option e

Step-by-step explanation:

The question asks us to determine the weight that should be advertised on a soap powder box so that only 3% of the boxes are underweight. Since the weight of the soap powder in the boxes follows a normal distribution with a mean of 33 ounces and a standard deviation of 0.8 ounces, we are looking for the z-score that corresponds to the 3rd percentile.

This is a problem associated with normal distribution and would require the use of z-tables or statistical software to find the value at which 3% of the values fall below it. The z-score for the 3rd percentile is approximately -1.88. Using the formula X = μ + zσ, where X is the weight to be advertised, μ is the mean, z is the z-score, and σ is the standard deviation, we can find the weight that should be advertised.

Plugging the values in, we get X = 33 + (-1.88)(0.8) = 32.49 ounces. Since we want to ensure only 3% are underweight, we should round up, making the correct advertised weight 32.6 ounces, which corresponds to option (e).