Question

Math

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 $(68-95-99.7 \%)$ 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 $x = 14.6$ using the formula $z = \frac{x - \mu}{\sigma}$ where $\mu = 13.1$ and $\sigma = 1.5$

step 2

Substitute the given values into the z-score formula: $z = \frac{14.6 - 13.1}{1.5}$

step 3

Compute the z-score: $z = \frac{1.5}{1.5} = 1$

step 4

Use the empirical rule to estimate the probability. Since $z = 1$ 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: $P(x < 14.6) \approx 50\% + \frac{68\%}{2} = 50\% + 34\% = 84\%$

Answer

The estimated probability of a meerkat living less than 14.6 years is approximately 84%.

Key Concept

Empirical Rule and Z-Score

Explanation

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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