Answer: the probability that the mean price for the sample was between $3.781 and $3.811 that week is 0.94122
Step-by-step explanation:
Given that;
sample size n = 32
mean μ = $3.796
standard deviation σ = 0.045
P(3.781 < x" < 3.811) = ?
Standard Error S.E = σ/√n = 0.045/√32 = 0.007955
z value for 3.781, z = x-μ/S.E = (3.781-3.796)/0.007955 = -1.8856 ≈ -1.89
z value for 3.811, z = x-μ/S.E = (3.811-3.796)/0.007955 = 1.8856 ≈ 1.89
P(3.781 < x" < 3.811) = P( -1.89 < z < 1.89)
= P(z < 1.89) - P(z < -1.89)
from z-score table
⇒ 0.9706 - 0.02938
⇒ 0.94122
Therefore the probability that the mean price for the sample was between $3.781 and $3.811 that week is 0.94122
