Questions: [2024 FILMLIT 1530 Introductory Statistics QO JBBIT DU Homework: S4 Sect 8.1: HW: Distribution of Sample Question Part of 6) The mean number of pregnancies is approximately normal with mean μ = 2.86 and standard deviation σ = 0.035. (d) What is the probability that a random sample of 258 pregnancies has a mean greater than 2.863? (c) If 100 independent random samples of size n = 258 are obtained from this population, we would expect about how many of those samples to have a sample mean greater than 2.863? (b) What is the probability that a random sample of 258 pregnancies has a mean less than 2.863? (a) What is the probability that a random sample of 258 pregnancies has a mean exactly equal to 2.863? (e) Round to the nearest integer, a sample of size n = 14 pregnancies has a mean number of pregnancies that is greater than 2.86 days is approximately: Help me solve this View an example Get more help Clear all Chat answer 11:51 PM]

[2024 FILMLIT 1530 Introductory Statistics QO JBBIT DU Homework: S4 Sect 8.1: HW: Distribution of Sample Question Part of 6) The mean number of pregnancies is approximately normal with mean μ = 2.86 and standard deviation σ = 0.035. (d) What is the probability that a random sample of 258 pregnancies has a mean greater than 2.863? (c) If 100 independent random samples of size n = 258 are obtained from this population, we would expect about how many of those samples to have a sample mean greater than 2.863? (b) What is the probability that a random sample of 258 pregnancies has a mean less than 2.863? (a) What is the probability that a random sample of 258 pregnancies has a mean exactly equal to 2.863? (e) Round to the nearest integer, a sample of size n = 14 pregnancies has a mean number of pregnancies that is greater than 2.86 days is approximately: Help me solve this View an example Get more help Clear all Chat answer 11:51 PM]
Transcript text: [2024 FILMLIT 1530 Introductory Statistics QO JBBIT DU Homework: S4 Sect 8.1: HW: Distribution of Sample Question Part of 6) The mean number of pregnancies is approximately normal with mean μ = 2.86 and standard deviation σ = 0.035. (d) What is the probability that a random sample of 258 pregnancies has a mean greater than 2.863? (c) If 100 independent random samples of size n = 258 are obtained from this population, we would expect about how many of those samples to have a sample mean greater than 2.863? (b) What is the probability that a random sample of 258 pregnancies has a mean less than 2.863? (a) What is the probability that a random sample of 258 pregnancies has a mean exactly equal to 2.863? (e) Round to the nearest integer, a sample of size n = 14 pregnancies has a mean number of pregnancies that is greater than 2.86 days is approximately: Help me solve this View an example Get more help Clear all Chat answer 11:51 PM]
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Solution

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Solution Steps

Step 1: Probability of Sample Mean Exactly Equal to 2.863

For a continuous distribution, the probability of a single point is given by:

\[ P(X = 2.863) = 0 \]

Thus, the probability that the sample mean is exactly equal to \(2.863\) is:

\[ \text{(a) } P = 0 \]

Step 2: Probability of Sample Mean Less Than 2.863

To find the probability that a random sample of \(n = 258\) pregnancies has a mean less than \(2.863\), we calculate:

\[ P(X < 2.863) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.3768) - \Phi(-\infty) = 0.9157 \]

Thus, the probability that the sample mean is less than \(2.863\) is:

\[ \text{(b) } P = 0.9157 \]

Step 3: Probability of Sample Mean Greater Than 2.863

To find the probability that a random sample of \(n = 258\) pregnancies has a mean greater than \(2.863\), we calculate:

\[ P(X > 2.863) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(1.3768) = 0.0843 \]

Thus, the probability that the sample mean is greater than \(2.863\) is:

\[ \text{(d) } P = 0.0843 \]

Step 4: Expected Number of Samples with Mean Greater Than 2.863

If \(100\) independent random samples of size \(n = 258\) are obtained from this population, the expected number of those samples to have a sample mean greater than \(2.863\) is calculated as:

\[ \text{Expected Samples} = 100 \times P(X > 2.863) = 100 \times 0.0843 = 8.43 \]

Thus, the expected number of samples with a mean greater than \(2.863\) is:

\[ \text{(c) } \text{Expected Samples} = 8.43 \]

Final Answer

\[ \text{(a) } P = 0 \\ \text{(b) } P = 0.9157 \\ \text{(c) } \text{Expected Samples} = 8.43 \\ \text{(d) } P = 0.0843 \]

\(\boxed{(a) = 0, (b) = 0.9157, (c) = 8.43, (d) = 0.0843}\)

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