Question

A population of values has a normal distribution with μ=242.8μ=242.8 and σ=66.4σ=66.4. You intend to draw a random sample of size n=249n=249.

Find the probability that a single randomly selected value is greater than 236.9.
P(X > 236.9) =

Find the probability that a sample of size n=249is randomly selected with a mean greater than 236.9.
P(M > 236.9) =

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

a)
Here, μ = 242.8, σ = 66.4 and x = 236.9. We need to compute P(X >= 236.9). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (236.9 - 242.8)/66.4 = -0.089

Therefore,
P(X >= 236.9) = P(z <= (236.9 - 242.8)/66.4)
= P(z >= -0.089)
= 1 - 0.4645 = 0.5355

b)
Here, μ = 242.8, σ = 66.4/sqrt(249) = 4.2079 and x = 236.9. We need to compute P(X >= 236.9). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (236.9 - 242.8)/4.2079 = -1.402

Therefore,
P(X >= 236.9) = P(z <= (236.9 - 242.8)/4.2079)
= P(z >= -1.402)
= 1 - 0.0805
= 0.9195