At Sunshine Lanes, each bowler pays $7 per game including shoe rental. Martin graphs the relationship between the number of game
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The probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328
<h3>How to determine the probability?</h3>The given parameters are:
- Sample size, n = 100
- Population proportion, p = 82%
Start by calculating the mean:
Calculate the standard deviation:
Within ± 0.02 of the population proportion are:
Calculate the z-scores at these points using:
So, we have:
The probability is then represented as:
P(x ± 0.02) = P(-0.43 < z < 0.43)
Using the z table of probabilities, we have:
P(x ± 0.02) = 0.3328
Hence, the probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328
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