The probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
цр = р
The standard deviation of this sampling distribution of sample proportion is:
бр = √ρ(1-ρ)÷n
The information provided is:
ρ = 0.22
ⁿ = 276
As the sample size is large, i.e. n = 276 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion.
Compute the value of P(р-p<0.06) as follows:
P(р-p<0.06) = P(р-p ÷ бp<0.06 ÷√0.22(1 - 0.22) ÷ 276
= P ( Z < 2.41 )
= 0.99202
≈ 0.992
Thus, the probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.
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