Question

The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives 20.5 years; the standard deviation is 2.5

σ=2.5 years.
Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between 16.6 and 24.4 years.

Answer 1

The probability of a zebra living between 16.6 and 24.4 years is approximately 88.12%.

Step-by-step explanation:

The question is asking to use the empirical rule to estimate the probability of a zebra living between 16.6 and 24.4 years. According to the empirical rule, approximately 68% of the values lie within one standard deviation of the mean, approximately 95% lie within two standard deviations, and approximately 99.7% lie within three standard deviations.

In this case, the mean lifespan of zebras is 20.5 years with a standard deviation of 2.5 years.

To estimate the probability of a zebra living between 16.6 and 24.4 years, we need to find the z-scores corresponding to these values and use the z-table to find the corresponding probabilities.

The z-score formula is:

z = (x - μ) / σ

where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation.

Calculating the z-scores:

z1 = (16.6 - 20.5) / 2.5 = -1.56

z2 = (24.4 - 20.5) / 2.5 = 1.56

Now we can use the z-table to find the probabilities:

P(z < -1.56) + P(-1.56 < z < 1.56) + P(z > 1.56)

Using the z-table, we can find that P(z < -1.56) is approximately 0.0594 and P(z > 1.56) is also approximately 0.0594.

Adding these probabilities together, we get:

P(-1.56 < z < 1.56) ≈ 1 - (P(z < -1.56) + P(z > 1.56)) ≈ 1 - (0.0594 + 0.0594) ≈ 0.8812

Therefore, the probability of a zebra living between 16.6 and 24.4 years is approximately 0.8812 or 88.12%.