Using The Information In A And B, Find The Probability

Using the information in A and B , find the probability that the difference in mean BMI for 30 women and 35 men selected independently and at random from the respective populations will exceed 3.20

Explain the problem: Given data from two studies, we need to find the probability that the difference in mean BMI between 30 women and 35 men exceeds 3.20. The data includes population means, standard deviations, and sample sizes for women and men aged 60 and over with normal skeletal muscle index.

Step 1: Summarize the provided information

For women (Study A):

  • Population mean BMI (μw): 23.1
  • Standard deviation (σw): 3.7
  • Sample size (nw): 30

For men (Study B):

  • Population mean BMI (μm): 24.7
  • Standard deviation (σm): 3.3
  • Sample size (nm): 35

Step 2: Define the random variables

Let w be the sample mean BMI of women, and m be the sample mean BMI of men. The difference in sample means is D = X̄w - X̄m.

Step 3: Find the distribution of the difference in sample means

Since both samples are independent, the distribution of the difference in sample means, D, is normal with mean:

μD = μw - μm = 23.1 - 24.7 = -1.6

Variance of D:

σ2D = (σw2 / nw) + (σm2 / nm)

Calculate:

  • (3.7)2 / 30 = 13.69 / 30 ≈ 0.4563
  • (3.3)2 / 35 = 10.89 / 35 ≈ 0.3111

Total variance: 0.4563 + 0.3111 ≈ 0.7674

Standard deviation:

σD = √0.7674 ≈ 0.876

Step 4: Find the probability that the difference exceeds 3.20

We want:

P(D > 3.20)

Standardize using the Z-score:

Z = (D - μD) / σD = (3.20 - (-1.6)) / 0.876 = (3.20 + 1.6) / 0.876 = 4.80 / 0.876 ≈ 5.48

Using standard normal tables or calculator:

P(Z > 5.48) is practically zero, as Z=5.48 is far in the tail of the normal distribution.

Therefore, the probability that the difference in mean BMI exceeds 3.20 is effectively zero, approximately 0.

Conclusion

Given the large Z-score, the probability that the difference in mean BMI between the two samples exceeds 3.20 is negligible, indicating that such a difference is extremely unlikely based on the given data.

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