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Given: p = 0.2, q = 1− p = 0.8,n = 100
Taking normal approximation to binomial, the sampling distribution of sample
proportion follows normal distribution.
Shape: Normal
p̂ ∼ N(µ,σ2), where
mean, µp̂ = p = 0.2
µp̂ = 0.2
variance, σ2 = p× (1− p)
n
= 0.2× 0.8100 = 0.0016
standard deviation, σp̂ =
√
σ2 =
√
0.2× 0.8
100 = 0.04
σp̂ = 0.04
p̂ ∼ N(0.2, 0.042)
A P (0.15 < p̂ < 0.25)
Normal Distribution, µ = 0.2, σ = 0.04, we convert this to standard normal
using z = p̂− µ
σ
z1 =
0.15− (0.2)
0.04 = −1.25
z2 =
0.25− (0.2)
0.04 = 1.25
P (−1.25 < Z < 1.25) = Area in between −1.25 and 1.25
Z
0.1056
1.25
−1.25
0
0.8944
0.8944− 0.1056 = 0.7888
P (0.15 < p̂ < 0.25) = P (−1.25 < Z < 1.25)
= P (Z < 1.25)− P (Z < −1.25) = 0.8944− 0.1056= 0.7888
P (Z < 1.25): in a z-table having area to the left of z, locate 1.2 in the left most column.
Move across the row to the right under column 0.05 and get value 0.8944
P (Z < −1.25): in a z-table having area to the left of z, locate -1.2 in the...