Suppose a large consignment of cameras contained $12 \%$ defectives.
If a sample of size 202 is selected, what is the probability that the sample proportion will be less than $9 \%$ ? Round your answer.to four deicimal places.

Step 1 ：We are given a problem of sampling distribution. The Central Limit Theorem can be used to solve this problem. The theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 2 ：In this problem, the population proportion (p) is given as 12% or 0.12. We are asked to find the probability that the sample proportion is less than 9% or 0.09. The sample size (n) is 202.

Step 3 ：The mean of the sampling distribution (μ) is equal to the population proportion (p). The standard deviation of the sampling distribution (σ) is calculated using the formula \(\sqrt{p(1-p)/n}\).

Step 4 ：We can standardize the sample proportion (0.09) using the formula z = (x - μ) / σ, where x is the sample proportion, μ is the mean of the sampling distribution, and σ is the standard deviation of the sampling distribution.

Step 5 ：Then we can use the standard normal distribution (z-distribution) to find the probability that the z-score is less than the calculated value.

Step 6 ：Let's calculate the values. The population proportion p = 0.12, the sample size n = 202, and the sample proportion x = 0.09. The mean of the sampling distribution μ = 0.12, and the standard deviation of the sampling distribution σ = 0.022864213899185835.

Step 7 ：Using the formula for z, we get z = -1.312094093078278.

Step 8 ：Using the z-distribution, we find that the probability is 0.0947.

Step 9 ：Final Answer: The probability that the sample proportion will be less than 9% is \(\boxed{0.0947}\).