Question

5.4.1 Question Help A population has a mean μ-84 and a standard deviation σ-36. Find the mean and standard deviation of a sampling distribution of sample means with sample size n 81. μί-d (simplify your answer.)

Answer 1

Solution:-

5.4.1) $\mu _{\bar{x}}=84$

5.4.18) The probability of a sample mean being less than 12748 or greater than 12751 is 0.5573.

Mean = 12748, S.D = 1.9

x₁ = 12748

x₂ = 12751

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z₁ = 0

z₂ = 1.5789

P( 0 > z < 1.58) = P(z < 0) + P(z > 1.58)

P( 0 > z < 1.58) = 0.50 + 0.0573

P( 0 > z < 1.58) = 0.5573