Question

A consulting firm believes that the mean number televisions in a household in the US is 2.4 with a standard deviation of 1.3 random sample of 40 households was obtained in the mean number of televisions per household in the sample was 2.78 what is the probability of getting a sample mean of 2.78 or greater when surveying 40 households

Answer 1

The probability of getting a sample mean of 2.78 or greater when surveying 40 households is approximately 0.053.

Step-by-step explanation:

To calculate the probability of getting a sample mean of 2.78 or greater when surveying 40 households, we need to use the sampling distribution of the sample mean. Due to the Central Limit Theorem, we know that the sampling distribution of the sample mean follows a normal distribution with a mean equal to the population mean (2.4) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (1.3/sqrt(40)).

To find the probability, we need to calculate the z-score for the sample mean of 2.78 using the formula: z = (x - mean) / (standard deviation).

Substituting the values: z = (2.78 - 2.4) / (1.3/sqrt(40)) = 2.78 - 2.4 / (1.3 * 0.180) = 0.38 / 0.234 = 1.624.

Next, we can use a standard normal distribution table or a calculator to find the probability of getting a z-score of 1.624 or greater. A standard normal distribution table gives us a probability of approximately 0.053.