Question 8 of 10, Step 1 of 1
Consider the probability that fewer than 11 out of 150 houses will lose power once a year. Assume the probability that a given house will lose power once a year is $9 \%$
Approximate the probability using the normal distribution. Round your answer to four decimal places.
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Step 1 ：This problem involves a binomial distribution, but we are asked to approximate it using the normal distribution. The normal approximation to the binomial distribution can be used when the number of trials is large.

Step 2 ：The mean (\(\mu\)) of a binomial distribution is np and the standard deviation (\(\sigma\)) is \(\sqrt{np(1-p)}\), where n is the number of trials and p is the probability of success.

Step 3 ：In this case, n = 150 (the number of houses) and p = 0.09 (the probability that a given house will lose power once a year). So, \(\mu = np = 150 \times 0.09 = 13.5\) and \(\sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.09 \times (1-0.09)} = 3.505\).

Step 4 ：We want to find the probability that fewer than 11 houses will lose power, so we want P(X < 11). When using the normal approximation, we need to apply a continuity correction, so we actually want to find P(X < 10.5).

Step 5 ：We can standardize this to the standard normal distribution by subtracting the mean and dividing by the standard deviation, to find P(Z < (10.5 - \(\mu\)) / \(\sigma\)).

Step 6 ：Substituting the values, we get Z = (10.5 - 13.5) / 3.505 = -0.856.

Step 7 ：Finally, we can use the cumulative distribution function (CDF) of the standard normal distribution to find this probability. The CDF at Z = -0.856 is approximately 0.1960.

Step 8 ：Final Answer: The approximate probability that fewer than 11 out of 150 houses will lose power once a year, using the normal distribution, is \(\boxed{0.1960}\).