Probability Help!

Question

The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 𝜇𝑀=188μM=188 milligrams per deciliter (mg/dl) and standard deviation 𝜎𝑀=41σM=41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 𝜇𝐵=170μB=170 mg/dl and standard deviation 𝜎𝐵=30σB=30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels 𝑥¯𝑀x¯M and 𝑥¯𝐵x¯B.

What is the probability of getting a difference in sample means 𝑥¯𝑀−𝑥¯𝐵x¯M−x¯B that’s less than 0 mg/dl?

Answer 1

You are given normal distributions for A and B and their standard deviations: difference of means 2-sample z test.

E(x_A,avg-x_B,avg) = μ_A-μ_B

σ_Diff = (σ₁²/n₁ + σ₁²/n₂)^1/2

The z statistic is (0 - (188 - 170))/σ_diff

Find Φ(z) = probability that the difference will be less that 0 for the sample. Φ(z) is the left area under the standard normal distribution function (The cum. distribution)

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