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Sagot :
The probability that the sample proportion will be between 4% and 8% is 0.9154 or 91.54%
According to the given information, the mean of the population µ = 0.05 and the sample size n = 900.
The standard deviation σ = √(p(1-p)/n = √(0.05(1-0.05)/900 = 0.0072648
In order to determine the probability that sample proportion will be between 4% and 8%, the standard score (z-score) must be calculated.
Z = (x - µ)/ (σ/√n)
Z (x = 0.04) = (0.04 – 0.05)/ 0.0072648 = -1.376
Z (x = 0.08) = (0.08 – 0.05)/ 0.0072648 = 4.129
Hence,
P(-1.376 < Z < 4.129) = P(z < 4.129) – P(z < 1.376)
From the table,
P(z < 4.129) = 0.9998
P(z <1.376) = 0.0844
Therefore,
P(z < 4.129) – P(z < 1.376) = 0.9998 – 0.0844 = 0.9154 or 91.54%
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