The Lengths Of Pregnancies In Days Are Normally Div

The Lengths Of Pregnancies In Number Of Days Are Normally Distributed

The lengths of pregnancies in number of days are normally distributed with a mean of 268 days and a standard deviation of 15 days. Answer the following questions and show your work. a. If a pregnancy length is randomly taken from this distribution, what is the probability that the length of that pregnancy is less than 260 days? b. If a sample of 25 pregnancies is taken from this distribution, what is the probability that their mean will be less than 260 days? c. Explain why these two answers (from a and b) are different.

Paper For Above instruction

The distribution of pregnancy lengths in days is an important aspect of obstetric health surveillance and planning. Given a normal distribution with a mean (μ) of 268 days and a standard deviation (σ) of 15 days, we can compute the probabilities for individual pregnancy lengths and sample means, which provides insights into typical gestation periods and their variability.

Part a: Probability that a randomly selected pregnancy is less than 260 days

To determine the probability that a pregnancy lasts less than 260 days, we first need to calculate the z-score corresponding to 260 days. The z-score is calculated as:

z = (X - μ) / σ = (260 - 268) / 15 = -8 / 15 ≈ -0.5333

Using standard normal distribution tables or statistical software, we find the probability corresponding to z ≈ -0.5333. The cumulative probability for z = -0.5333 is approximately 0.297, indicating that about 29.7% of pregnancies are expected to be shorter than 260 days.

Therefore, the probability that a randomly selected pregnancy lasts less than 260 days is approximately 0.297 or 29.7%.

Part b: Probability that the mean of a sample of 25 pregnancies is less than 260 days

When considering a sample of size n = 25, the sampling distribution of the sample mean (x̄) will also be normal with the same mean μ = 268 days, but with a standard error (SE) calculated as:

SE = σ / √n = 15 / √25 = 15 / 5 = 3

The z-score for the sample mean being less than 260 days is then:

z = (X̄ - μ) / SE = (260 - 268) / 3 = -8 / 3 ≈ -2.6667

Consulting the standard normal distribution table or software for z ≈ -2.6667 yields a cumulative probability of approximately 0.0038, meaning there's only about a 0.38% chance that the average length of a randomly chosen sample of 25 pregnancies is less than 260 days.

Part c: Explanation of the difference between the two probabilities

The probabilities obtained in parts a and b are different due to the concepts of sampling distribution and variability. In part a, the probability pertains to a single pregnancy length, which directly follows the population distribution. This individual probability is relatively high because the value of 260 days is only about half a standard deviation below the mean.

In contrast, part b addresses the probability that the average of a sample of size 25 is less than 260 days. When sampling, the variability of the sampling distribution of the mean decreases as the sample size increases, following the standard error formula. Since the standard error (3 days) is much smaller than the population standard deviation, the distribution of the sample mean is more concentrated around the population mean.

Consequently, the likelihood of observing a sample mean as extreme as 260 days (which is almost 2.667 standard errors below the population mean) is very low, leading to a much smaller probability (~0.0038) compared to the individual probability (~0.297). This illustrates the principle that larger samples tend to produce more precise estimates, reducing variability in the sampling distribution of the mean.

Additional Analysis: Organic Certification of Growers

In a separate context, consider the analysis where 494 out of 845 coffee growers from southern Mexico are certified organic. To determine whether fewer than 60% of these growers are certified, a hypothesis test was conducted. The null hypothesis (H₀) states that p = 0.60, while the alternative hypothesis (H₁) suggests p

Calculations yielded a p-value of approximately 0.1867. This p-value exceeds the common significance levels of 0.05 and 0.10, indicating insufficient evidence to reject the null hypothesis. Despite observing a sample proportion of around 58.4% (494/845 ≈ 0.584), statistically, we cannot conclude that the true proportion is less than 60% at the 5% significance level.

The margin of error associated with this hypothesis test provides a measure of precision, which in this context is approximately 0.037. This margin reflects the range within which the true population proportion likely falls, given a 95% confidence level.

Conclusion

Understanding the differences in probabilities between individual observations and sample means is crucial in statistical inference. The probability that a single pregnancy is less than 260 days reflects the inherent variability in gestation lengths, whereas the probability for the sample mean accounts for the decreased variability due to larger sample size, resulting in more precise estimates. This distinction underscores the importance of sample size in statistical analysis and inference.

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