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Posted 3 months ago

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 (689599.7%)(68-95-99.7 \%) to estimate the probability of a meerkat living less than 14.6 years.
\square %\%
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Calculate the z-score for x=14.6x = 14.6 using the formula z=xμσz = \frac{x - \mu}{\sigma} where μ=13.1\mu = 13.1 and σ=1.5\sigma = 1.5
step 2
Substitute the given values into the z-score formula: z=14.613.11.5z = \frac{14.6 - 13.1}{1.5}
step 3
Compute the z-score: z=1.51.5=1z = \frac{1.5}{1.5} = 1
step 4
Use the empirical rule to estimate the probability. Since z=1z = 1 corresponds to 1 standard deviation from the mean, approximately 68% of the data falls within 1 standard deviation on either side of the mean
step 5
To find the probability of a meerkat living less than 14.6 years, we take half of 68% (because we are only looking at one side of the distribution) and add it to the 50% of the data that falls below the mean
step 6
Calculate the probability: P(x<14.6)50%+68%2=50%+34%=84%P(x < 14.6) \approx 50\% + \frac{68\%}{2} = 50\% + 34\% = 84\%
The estimated probability of a meerkat living less than 14.6 years is approximately 84%.
Key Concept
Empirical Rule and Z-Score
The empirical rule is used to estimate the probability that a data point falls within a certain number of standard deviations from the mean in a normal distribution. The z-score helps to determine the position of a value within the distribution.

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