The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 13.1 years; the standard deviation is 1.5 years.

Use the empirical rule $(68-95-99.7 \%)$ to estimate the probability of a meerkat living less than 14.6 years.
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Question

The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 13.1 years; the standard deviation is 1.5 years.

Use the empirical rule $(68-95-99.7 \%)$ to estimate the probability of a meerkat living less than 14.6 years.
$\square$ $\%$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the mean $\mu$ and standard deviation $\sigma$ of the normal distribution for meerkat lifespans. ‖ ‖ step 2 ⋮ Calculate the z-score for the value 14.6 years using the formula $z = \frac{x - \mu}{\sigma}$. ‖ ‖ step 3 ⋮ Use the empirical rule to estimate the probability of a meerkat living less than 14.6 years. ‖ ‖ step 4 ⋮ Since 14.6 is within one standard deviation above the mean, approximately 68% of the data lies within one standard deviation of the mean. ‖ ‖ step 5 ⋮ To find the probability of a meerkat living less than 14.6 years, we take half of 68% (which accounts for the lower side of the mean) and add it to the 50% (which accounts for the lower half of the distribution). ‖ ‖ step 6 ⋮ The calculation is $50\% + \frac{68\%}{2} = 50\% + 34\% = 84\%$. ‖ ∻[1] Answer∻ ⚹ 84% ⚹ ∻Key Concept∻ ⚹Empirical Rule⚹ ∻Explanation∻ ⚹The empirical rule states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. To find the probability of a value being less than a certain number within one standard deviation above the mean, we add half of 68% to 50%.⚹◊What is the formula for calculating the z-score when using the empirical rule in a normal distribution scenario?⍭ Generate me a similar question◊

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