- The length of pregnancy for the Asian elephant has an approximately normal distribution with an average length of 609 days and standard deviation of 31 days.
A) What percent of the elephant pregnancies last between 600 and 615 days?
B) What percent of the elephant pregnancies last fewer than 595 days?
C) What percent of the elephant pregnancies last more than 620 days?
D) How long do the longest 5% of all elephant pregnancies last?
E) The shortest 20% of all elephant pregnancies last fewer than how many days?
F) The middle 50% of all elephant pregnancies fall between how many days?
Let’s analyze the problem step-by-step using properties of the normal distribution.
Given:
- Mean (μ) = 609 days
- Standard deviation (σ) = 31 days
- Distribution: Approximately normal
A) What percent last between 600 and 615 days?
Step 1: Convert 600 and 615 to Z-scores
$$
Z = \frac{X – \mu}{\sigma}
$$
- For 600 days:
$$
Z_{600} = \frac{600 – 609}{31} = \frac{-9}{31} \approx -0.29
$$
- For 615 days:
$$
Z_{615} = \frac{615 – 609}{31} = \frac{6}{31} \approx 0.19
$$
Step 2: Use standard normal tables or a calculator
- $P(Z < 0.19) \approx 0.5753$ (57.53%)
- $P(Z < -0.29) \approx 0.3859$ (38.59%)
Step 3: Find the probability between these Z-values
$$
P(-0.29 < Z < 0.19) = P(Z < 0.19) – P(Z < -0.29) = 0.5753 – 0.3859 = 0.1894
$$
Answer: Approximately 18.94% of pregnancies last between 600 and 615 days.
B) What percent last fewer than 595 days?
Step 1: Convert 595 to Z
$$
Z_{595} = \frac{595 – 609}{31} = \frac{-14}{31} \approx -0.45
$$
Step 2: Look up $P(Z < -0.45)$
$$
P(Z < -0.45) \approx 0.3264
$$
Answer: Approximately 32.64% last fewer than 595 days.
C) What percent last more than 620 days?
Step 1: Convert 620 to Z
$$
Z_{620} = \frac{620 – 609}{31} = \frac{11}{31} \approx 0.35
$$
Step 2: Find $P(Z > 0.35) = 1 – P(Z < 0.35)$
$$
P(Z < 0.35) \approx 0.6368
$$
So,
$$
P(Z > 0.35) = 1 – 0.6368 = 0.3632
$$
Answer: Approximately 36.32% last more than 620 days.
D) How long do the longest 5% last?
We want the 95th percentile $P_{95}$.
Step 1: Find $Z$ for 95th percentile
From the Z-table, $Z_{0.95} \approx 1.645$.
Step 2: Convert back to days
$$
X = \mu + Z \times \sigma = 609 + 1.645 \times 31 = 609 + 50.995 = 659.995 \approx 660 \text{ days}
$$
Answer: The longest 5% last more than approximately 660 days.
E) The shortest 20% last fewer than how many days?
We want the 20th percentile $P_{20}$.
Step 1: Find $Z$ for 20th percentile
From the Z-table, $Z_{0.20} \approx -0.84$.
Step 2: Convert back to days
$$
X = 609 + (-0.84) \times 31 = 609 – 26.04 = 582.96 \approx 583 \text{ days}
$$
Answer: The shortest 20% last fewer than approximately 583 days.
F) The middle 50% fall between how many days?
The middle 50% corresponds to the interquartile range (IQR), between the 25th percentile $P_{25}$ and the 75th percentile $P_{75}$.
Step 1: Find $Z$ for 25th and 75th percentiles
- $Z_{0.25} \approx -0.674$
- $Z_{0.75} \approx 0.674$
Step 2: Convert to days
- Lower bound:
$$
X_{25} = 609 + (-0.674) \times 31 = 609 – 20.89 = 588.11 \approx 588 \text{ days}
$$
- Upper bound:
$$
X_{75} = 609 + 0.674 \times 31 = 609 + 20.89 = 629.89 \approx 630 \text{ days}
$$
Answer: The middle 50% last between approximately 588 and 630 days.
Detailed Explanation (300+ words):
The problem involves a normal distribution describing the length of Asian elephant pregnancies, with a mean (average) of 609 days and a standard deviation of 31 days. Because the distribution is normal, the empirical rule and standard normal (Z) scores allow us to calculate percentages and specific values related to different parts of the distribution.
Part A asks for the percentage between 600 and 615 days. By converting these to Z-scores, which represent how many standard deviations these values are from the mean, we can use standard normal distribution tables to find probabilities. The difference between the cumulative probabilities at these two Z-values gives the percentage between them.
Part B asks for the percentage lasting fewer than 595 days, which requires finding the cumulative probability up to the Z-score for 595. This gives us the area under the normal curve to the left of that value.
Part C is similar but asks for pregnancies longer than 620 days. We find the cumulative probability up to 620 days and subtract from 1 to get the tail area on the right.
For Part D, we’re asked for the pregnancy length marking the longest 5%, which is the 95th percentile. We find the Z-score corresponding to the 95th percentile and convert it back to days using the mean and standard deviation. This length represents the cutoff after which only 5% of pregnancies are longer.
Part E focuses on the shortest 20%, i.e., the 20th percentile. The process is the same but for the lower tail of the distribution.
Part F looks for the middle 50%, which is the interquartile range between the 25th and 75th percentiles. Using the Z-scores for these percentiles and converting back gives the range of days where half of all pregnancies fall.
Overall, understanding Z-scores, percentiles, and using standard normal distribution tables are crucial for interpreting and solving these problems involving normally distributed data.
If you want, I can also provide detailed calculation steps or visualizations!
