Answer:
0.912
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 182 mg/dL
Standard deviation = σ = 37 mg/dL
So, for a sample of 1000 men
The sample mean = μₓ = population mean = μ = 182 mg/dL
The standard deviation of the sample mean = σₓ = (σ/√n)
where σ = population standard deviation and n = sample size
σₓ = (37/√1000) = 1.17 mg/dL
The variation in standard deviation of the sample mean as the sample reduces as the sample size increases especially with large numbers such as 1000.
But now, we require the probability that the sample mean falls within ±2 mg/dL of the population mean. P(|x - μ| ≤ 2)
We first standardize this value.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (±2)/1.17 = ± 1.71
P(|x - μ| ≤ 2) = P(|z| ≤ 1.71) = P(-1.71 ≤ z ≤ 1.71)
= P(z ≤ 1.71) - P(z ≤ -1.71)
= 0.95637 - 0.04363 = 0.9124 = 0.912 to 3 d.p
Hope this Helps!!!
