Suppose a large shipment of microwave ovens contained 18% defectives. If a sample of size 228 is selected, what isthe probability that the sample proportion will be greater than 14%? (Round your answer to 4 decimal places)
To find the probability that the sample proportion will be greater than 14%, we can use the normal distribution approximation for sample proportions. This approximation holds when the sample size is large enough and the population distribution is approximately normal.
To calculate this probability, we need to find the mean and standard deviation of the sample proportion.
The mean of the sample proportion can be calculated as:
mean = proportion of defectives in the population = 18% = 0.18
The standard deviation of the sample proportion can be calculated as:
standard deviation = sqrt[(p*q)/n]
Where:
p = proportion of defectives in the population = 18% = 0.18
q = 1 - p = 1 - 0.18 = 0.82
n = sample size = 228
Substituting these values into the formula, we get:
standard deviation = sqrt[(0.18 * 0.82) / 228] ≈ 0.026
Now, we can use the standard normal distribution to calculate the probability.
The z-score can be calculated as:
z = (x - mean) / standard deviation
Where x is the value we want to find the probability for, which is 14% = 0.14 in this case.
Substituting the values, we get:
z = (0.14 - 0.18) / 0.026 ≈ -1.5385
Now, using the z-table or a statistical calculator, we can find the probability associated with this z-score or greater. In this case, we want to find the probability for z > -1.5385.
The probability can be calculated as:
probability = 1 - P(z <= -1.5385)
Using a z-table or a statistical calculator, we can find the value of P(z <= -1.5385) is approximately 0.0621.
Therefore, the probability that the sample proportion will be greater than 14% is:
probability = 1 - 0.0621 ≈ 0.9379
Rounding to 4 decimal places, the probability is approximately 0.9379.