Question

In a study researchers reported the mean BMI for men 60 and older to be 24.7 with a standard deviation of 3.3 and the mean BMI for women 60 and older to be 23.1 with a standard deviation of 3.7. Using these values as the populations mean and standard deviation for men and women respectively, find the probability that the difference in the mean BMI (men-women) for 45 women and 50 men selected independently and at random will exceed 2.1.

Answer 1

Answer:

Probability that the difference in the mean BMI (men-women) for 45 women and 50 men selected independently and at random will exceed 2.1 = 0.2451

Step-by-step explanation:

The central limit theorem helps us to obtain the mean and the standard deviation of any sampling distribution.

Given that the sample was obtained from a normal distribution or an approximately normal distribution & it was obtained using random sampling techniques with each variable independent of one another and with each sample with adequate sample size,

Mean of sampling distribution (μₓ) = Population mean (μ)

Standard deviation of the sampling distribution = σₓ = (σ/√N)

where σ = population mean

N = Sample size

For the 45 women

μₓ = μ = 23.1

σₓ = (σ/√N) = (3.7/√45) = 0.552

For the 50 men

μₓ = μ = 24.7

σₓ = (σ/√N) = (3.3/√50) = 0.467

To find the probability that the difference in the mean BMI (men-women) for 45 women and 50 men selected independently and at random will exceed 2.1, we need to combine the distributions.

New distribution = (BMI of men) - (BMI of women) = x = X₁ - X₂

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

λ₁ = 1, λ₂ = -1

μ₁ = 24.7, μ₂ = 23.1

σ₁ = 0.467, σ₂ = 0.552

Combined mean = (Mean of men) - (Mean of women) = 24.7 - 23.1 = 1.6

Combined Variance = (1²×0.467²) + [(-1)²×(0.552²)] = 0.522793

Combined standard deviation = √0.522793 = 0.723

Probability that the difference in the mean BMI (men-women) for 45 women and 50 men selected independently and at random will exceed 2.1 = P(x > 2.1)

Note that the resulting distribution from the combination of distributions is still a normal distribution since the distributions combined were normal distributions too.

Hence, we first normalize or standardize 2.1

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (2.1 - 1.6)/0.723 = 0.69

To determine the required probability

P(x > 2.1) = P(z > 0.69)

We'll use data from the normal distribution table for these probabilities

P(x > 2.1) = P(z > 0.69) = 1 - P(z ≤ 0.69)

= 1 - 0.7549

= 0.2451

Hope this Helps!!!