Elephants have the longest pregnancy of all mammals. One species of elephant has a mean gestation period of 525 days and standard deviation of 32 days. These pregnancy lengths can be modeled by a normal distribution. What is the probability that a randomly selected elephant of this species has a pregnancy that lasts between 500 and 550 days? There is a 25% chance that an elephant from this species will have a pregnancy that lasts longer than __ days.
The Correct Answer and Explanation is:
We are given the following information about elephant gestation:
- Mean (μ) = 525 days
- Standard deviation (σ) = 32 days
- The distribution is normal
Part 1: Probability that a pregnancy lasts between 500 and 550 days
We want:
P(500 ≤ X ≤ 550)
To find this probability, we convert raw scores (X-values) into z-scores using: z=X−μσz = \frac{X – \mu}{\sigma}
- Z-score for 500 days:
z=500−52532=−2532≈−0.78z = \frac{500 – 525}{32} = \frac{-25}{32} \approx -0.78
- Z-score for 550 days:
z=550−52532=2532≈0.78z = \frac{550 – 525}{32} = \frac{25}{32} \approx 0.78
Using a standard normal distribution table (or calculator):
- P(Z ≤ 0.78) ≈ 0.7823
- P(Z ≤ -0.78) ≈ 0.2177
P(500≤X≤550)=P(Z≤0.78)−P(Z≤−0.78)=0.7823−0.2177=0.5646P(500 ≤ X ≤ 550) = P(Z ≤ 0.78) – P(Z ≤ -0.78) = 0.7823 – 0.2177 = 0.5646
✅ Answer: The probability is approximately 0.5646 or 56.46%.
Part 2: There is a 25% chance that a pregnancy lasts longer than how many days?
We are looking for the X-value where the upper tail of the normal distribution contains 25% of the values.
That means: P(X>x)=0.25⇒P(X≤x)=0.75P(X > x) = 0.25 \Rightarrow P(X ≤ x) = 0.75
Using a z-table, the z-score corresponding to 0.75 cumulative probability is approximately z = 0.674.
Now convert z back to an X-value using: X=μ+zσ=525+(0.674)(32)≈525+21.57=546.57X = \mu + z\sigma = 525 + (0.674)(32) \approx 525 + 21.57 = 546.57
✅ Answer: About 546.57 days
Summary
- Probability of a pregnancy lasting between 500 and 550 days is ~56.46%.
- There is a 25% chance a pregnancy lasts more than ~546.57 da