Math Problem Statement

Assume that the duration of human pregnancies can be described by a Normal model with mean 267 days and standard deviation 16 days. ​a) What percentage of pregnancies should last between 270 and 282 ​days? ​b) At least how many days should the longest 15​% of all pregnancies​ last? ​c) Suppose a certain obstetrician is currently providing prenatal care to 46 pregnant patients. Let y overbar represent the mean length of their pregnancies. According to the Central Limit​ Theorem, what is the distribution of this sample​ mean, y overbar​? Specify the​ model, mean, and standard deviation. ​d) What's the probability that the mean duration of these​ patients' pregnancies will be less than 259 ​days

Solution

Let's solve each part step by step:


Part (a): Percentage of pregnancies lasting between 270 and 282 days

We are tasked with finding the percentage of pregnancies lasting between 270 and 282 days. The model is a normal distribution with:

  • Mean (μ\mu) = 267 days
  • Standard deviation (σ\sigma) = 16 days

The corresponding zz-scores are calculated as: z=xμσz = \frac{x - \mu}{\sigma}

For x=270x = 270: z1=27026716=316=0.1875z_1 = \frac{270 - 267}{16} = \frac{3}{16} = 0.1875

For x=282x = 282: z2=28226716=1516=0.9375z_2 = \frac{282 - 267}{16} = \frac{15}{16} = 0.9375

Using standard normal distribution tables or a calculator, the cumulative probabilities are:

  • P(Z0.1875)0.5744P(Z \leq 0.1875) \approx 0.5744
  • P(Z0.9375)0.8255P(Z \leq 0.9375) \approx 0.8255

The percentage of pregnancies lasting between 270 and 282 days is: P(270X282)=P(Z0.9375)P(Z0.1875)=0.82550.5744=0.2511P(270 \leq X \leq 282) = P(Z \leq 0.9375) - P(Z \leq 0.1875) = 0.8255 - 0.5744 = 0.2511

Thus, 25.11% of pregnancies last between 270 and 282 days.


Part (b): Longest 15% of all pregnancies

We need to find the value of xx such that the top 15% of pregnancies have lengths greater than xx. The corresponding cumulative probability is: P(Zz)=10.15=0.85P(Z \leq z) = 1 - 0.15 = 0.85

From standard normal tables or a calculator, the zz-score corresponding to P(Zz)=0.85P(Z \leq z) = 0.85 is approximately: z=1.036z = 1.036

Converting this zz-score back to the original scale: x=μ+zσ=267+1.03616267+16.576=283.576x = \mu + z \cdot \sigma = 267 + 1.036 \cdot 16 \approx 267 + 16.576 = 283.576

Thus, the longest 15% of pregnancies last at least 283.6 days.


Part (c): Distribution of the sample mean (yˉ\bar{y})

According to the Central Limit Theorem, the sample mean (yˉ\bar{y}) is normally distributed with:

  • Mean: μyˉ=μ=267\mu_{\bar{y}} = \mu = 267 days
  • Standard deviation: σyˉ=σn=16462.36\sigma_{\bar{y}} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{46}} \approx 2.36

Thus, the distribution of yˉ\bar{y} is: yˉN(267,2.36)\bar{y} \sim N(267, 2.36)


Part (d): Probability that the mean duration is less than 259 days

We are tasked with finding P(yˉ<259)P(\bar{y} < 259), where:

  • μyˉ=267\mu_{\bar{y}} = 267
  • σyˉ2.36\sigma_{\bar{y}} \approx 2.36

The corresponding zz-score is: z=yˉμyˉσyˉ=2592672.36=82.363.39z = \frac{\bar{y} - \mu_{\bar{y}}}{\sigma_{\bar{y}}} = \frac{259 - 267}{2.36} = \frac{-8}{2.36} \approx -3.39

Using standard normal tables or a calculator: P(Z<3.39)0.00035P(Z < -3.39) \approx 0.00035

Thus, the probability that the mean duration of these patients' pregnancies is less than 259 days is approximately 0.035%.


Summary of Results

  1. (a) P(270X282)=25.11%P(270 \leq X \leq 282) = 25.11\%
  2. (b) Longest 15% of pregnancies last at least 283.6 days.
  3. (c) yˉN(267,2.36)\bar{y} \sim N(267, 2.36)
  4. (d) Probability P(yˉ<259)P(\bar{y} < 259) is approximately 0.035%.

Do you have any questions or want a more detailed explanation for any part?


Follow-Up Questions

  1. How is the zz-score used to convert between raw scores and probabilities in a normal distribution?
  2. Why does the Central Limit Theorem apply when dealing with sample means?
  3. Can you explain how the standard deviation of the sample mean decreases as the sample size increases?
  4. How would the results change if the sample size in part (c) were doubled?
  5. What are some real-world applications of normal distribution in medicine?

Tip

When using the Central Limit Theorem, always verify that the sample size is large enough for the theorem to apply effectively (usually n30n \geq 30).

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Probability
Central Limit Theorem
Sampling Distribution

Formulas

z = (x - μ) / σ
P(Z ≤ z) for cumulative probability
σ_sample = σ / √n

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12, Undergraduate Statistics