Answer :
Final answer:
A shopper at Browseworld's would need to spend about 41.8 minutes in the store to be in the longest 30% of shopping durations. This is based on a mean of 37.1 minutes and standard deviation of 9.1 minutes.
Explanation:
The question pertains to the topic of normal distribution in probability and statistics. Here, we have the duration of shoppers' time in browseworld's new retail outlets that is normally distributed with a mean of 37.1 minutes and a standard deviation of 9.1 minutes.
To find the duration that puts a shopper in the longest 30 percent, we need to find the 70th percentile of the normal distribution (since the longer duration accounts for the upper part of the distribution).
In a standard normal distribution table, the Z score is about 0.52 for the 70th percentile. To convert this Z score to our specific situation, we use the formula Z = (X - mean) / SD, where X is the value we're looking for.
Rearranging the formula to solve for X gives us X = Z * SD + mean. Substituting the values, we find X = 0.52 * 9.1 + 37.1. This calculation gives us X ≈ 41.8 minutes. So a visit must be about 41.8 minutes or longer to put a shopper in the longest 30 percent.
Learn more about Normal Distribution here:
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