Amanufacturing company measures the weight of boxes before shipping them to the customers. if the box weights have a population mean and standard deviation of 90 lb and 24 lb, respectively, then based on a sample size of 36 boxes, what is the probability that the average weight of the boxes will exceed 94 lb?

The probability that the average weight of the boxes will exceed 94 pounds  is 43.25%

Step-by-step explanation:

1. Let's review the information provided to answer the question correctly:

Mean of the population of box weights = 90 pounds

Standard deviation of the population of box weights = 24 pounds

Sample size = 36 boxes

2. What is the probability that the average weight of the boxes will exceed 94 lb?

For answering this question, we will use the z-scores table.

For using the correct z-score, we should use the correct standard deviation. In this case, we have that:

94 - 90 = 4 pounds above the mean.

4 pounds are 4/24 of the standard deviation,

therefore, simplifying:

4/24 = 1/6 = 0.1666and we'll round to 0.17

z-score is 0.17 because 4 pounds is 1/6 of the standard deviation (24).

The probability of a z-score is 0.5675, but in this case we're asked by a weight over 94 pounds, not under 94 pounds, thus:

1 - 0.5675 = 0.4325

The probability that the average weight of the boxes will exceed 94 pounds  is 43.25%

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step-by-step explanation:

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if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads