When sampling is done for the proportion of defective items in a large shipment, where the population proportion is 0.18 and the sample size is 200, what is the probability that the sample proportion will be at least 0.20
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To find the probability that the sample proportion will be at least 0.20, we can use the normal distribution approximation since the sample size (200) is large enough and satisfies the conditions for the Central Limit Theorem.
To calculate this, we'll first convert the problem into a standard normal distribution by standardizing the sample proportion using the formula:
z = (p̂ - p) / sqrt((p * (1 - p)) / n)
Where:
p̂ = sample proportion (0.20 in this case)
p = population proportion (0.18 in this case)
n = sample size (200 in this case)
Substituting the values into the formula, we have:
z = (0.20 - 0.18) / sqrt((0.18 * (1 - 0.18)) / 200)
= 0.02 / sqrt(0.1476 / 200)
= 0.02 / sqrt(0.000738)
≈ 2.151
Now that we have the standardized value (z-score), we can find the probability using a standard normal distribution table or a calculator. The probability of the sample proportion being at least 0.20 corresponds to the area under the normal distribution curve to the right of the z-score.
Looking up the z-score of 2.151 in a standard normal distribution table, we find that the corresponding area is approximately 0.0158 (or 1.58%).
Therefore, the probability that the sample proportion will be at least 0.20 is approximately 0.0158 (or 1.58%).