A sample of 51 observations will be taken from a process (an infinite population). The population proportion equals .85. The probability that the sample proportion will be between .9115 and .946 is _____.
A) .8633
B) .6900
C) .0819
D) .0345

Question

Quizplus · Accepted Answer

We can use the central limit theorem to approximate the sampling distribution of the sample proportion with a normal distribution. The mean of this normal distribution is equal to the population proportion (p) and the standard deviation is equal to the square root of [(p*(1-p))/n], where n is the sample size.

Plugging in the values given, we get:
Mean = p = 0.85
Standard deviation = sqrt[(p*(1-p))/n] = sqrt[(0.85*0.15)/51] = 0.064

To find the probability that the sample proportion is between 0.9115 and 0.946, we need to standardize these values by subtracting the mean and dividing by the standard deviation:
z1 = (0.9115 - 0.85) / 0.064 = 0.969
z2 = (0.946 - 0.85) / 0.064 = 1.500

Using a standard normal distribution table, we can find the probabilities associated with these z-values:
P(0.969 < Z < 1.500) = P(Z < 1.500) - P(Z < 0.969) = 0.9332 - 0.8513 = 0.0819

Therefore, the probability that the sample proportion will be between 0.9115 and 0.946 is 0.0819, which corresponds to answer choice C.

A Sample of 51 Observations Will Be Taken from a Process