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Assume that the height, X, of a college woman is a normally distributed random variable with a mean of 65 inches and a standard deviation of 3 inches. Suppose that we sample the heights of 180 randomly chosen college women. Let M be the sample mean of the

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Answer #1

To solve the given questions, we need to use the properties of the normal distribution. We'll use the mean (μ) and standard deviation (σ) provided:

a) Probability that X < 59: P(X < 59) = Φ((59 - μ) / σ) P(X < 59) = Φ((59 - 65) / 3) P(X < 59) = Φ(-2) ≈ 0.0228

b) Probability that X > 59: P(X > 59) = 1 - P(X < 59) P(X > 59) ≈ 1 - 0.0228 ≈ 0.9772

c) Probability that all 180 measurements are greater than 59: This probability is obtained by raising the probability of a single measurement being greater than 59 to the power of 180. P(all measurements > 59) = (P(X > 59))^180 ≈ 0.9772^180 ≈ 0

d) Expected value of S (sum of 180 height measurements): E(S) = n * μ = 180 * 65 = 11,700 inches

e) Standard deviation of S: σ(S) = √(n) * σ = √(180) * 3 ≈ 24.74 inches

f) Probability that S - 180 * 65 > 10: P(S - 180 * 65 > 10) = P((S - E(S)) / σ(S) > (10 / σ(S))) P(S - 180 * 65 > 10) = P((S - 11,700) / 24.74 > 10 / 24.74) P(S - 180 * 65 > 10) ≈ P(Z > 0.4034) ≈ 0.3439 (using standard normal distribution table)

g) Standard deviation of S - 180 * 65: σ(S - 180 * 65) = σ(S) = 24.74 inches

h) Expected value of M (sample mean of the 180 height measurements): E(M) = μ = 65 inches

i) Standard deviation of M: σ(M) = σ(X) / √(n) = 3 / √(180) ≈ 0.2236 inches

j) Probability that M > 65.41: P(M > 65.41) = 1 - P(M ≤ 65.41) = 1 - Φ((65.41 - μ) / (σ(X) / √n)) P(M > 65.41) = 1 - Φ((65.41 - 65) / (3 / √180)) P(M > 65.41) ≈ 1 - Φ(0.4837) ≈ 1 - 0.6877 ≈ 0.3123 (using standard normal distribution table)

k) Standard deviation of 180 * M: σ(180 * M) = 180 * σ(M) = 180 * 0.2236 ≈ 40.45 inches

l) If the probability of X > k is equal to 0.3, then what is k? P(X > k) = 0.3 Using the standard normal distribution table or a calculator, we can find the value of k where P(X > k) ≈ 0.3. The value of k is approximately 0.5244.

Note: In some questions, we used the standard normal distribution table or a calculator to find the probabilities associated with the standard normal distribution (Z-distribution). The standard normal distribution table provides probabilities for Z-scores (standard scores) from -3.49 to 3.49. For values outside this range, we can use a calculator or statistical software to find the probabilities.


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