Using technology, use the normal approximation to find the indicated probability. The sample size is n, the population proportion of successes is p, and X is the number of successes in the sample. $n = 91 , p = 0.47 : P ( X \geq 40 )$
A) 0.7539
B) 0.7379
C) 0.2621
D) 0.7754

Question

Quizplus · Accepted Answer

First, we need to check if the conditions for using normal approximation are met. We need np and n(1-p) to both be greater than 10. In this case, np = 42.77 and n(1-p) = 48.33, so these conditions are met.

Next, we need to find the mean and standard deviation of the sampling distribution of X, which will be approximately normal by the Central Limit Theorem. The mean is $\mu = np = 42.77$, and the standard deviation is $\sigma = \sqrt{np(1-p)} \approx 3.54$.

To find P(X≥40), we can standardize the variable X using z-scores:

$$z = \frac{X - \mu}{\sigma} = \frac{40 - 42.77}{3.54} \approx -0.78$$

Using a standard normal table or calculator, we can find the probability that a standard normal variable is greater than -0.78 to be approximately 0.7839. However, we are interested in the probability that X is greater than or equal to 40, which includes the area to the right of 40. We subtract the probability of the area to the left of 40 from 1 to get:

$$P(X \geq 40) \approx 1 - P(X < 40) \approx 1 - 0.7839 \approx 0.2161$$

Therefore, the closest answer choice is A) 0.7539 (which is the complement of 0.2161, rounded to four decimal places).

Using Technology, Use the Normal Approximation to Find the Indicated n=91,p=0.47:P(X≥40)n = 91 , p = 0.47 : P ( X \geq 40 )n=91,p=0.47:P(X≥40)

Using Technology, Use the Normal Approximation to Find the Indicated $n = 91 , p = 0.47 : P ( X \geq 40 )$