Question

The mean per capita income is 20,908 dollars per annum with a standard deviation of 407 dollars per annum. What is the probability that the sample mean would be less than 20948 dollars if a sample of 55 persons is randomly selected? Round your answer to four decimal places.

Answer 1

P ( x_sample mean < 20948 ) = 0.7673

Explanation:

Given:-

- The mean, u = 20,908

- Standard deviation, σ = 407

- Sample size, n = 55.

Find:-

The required probability is P ( x_sample mean < 20948 )

Solution:-

In this question, the probability of the sample mean is calculated by the z-test statistic under the normal probability. The z-test statistic is a normal test statistic and it is used for a large sample size in a normal probability problem.

- Determine the Z-score value for the statistics:

P ( x_sample mean < 20948 ) = P ( Z < (x - u )*√n / σ )

= P ( Z < (20,948 - 20,908 )*√55 / 407)

= P ( Z < 0.73)

- Using the Z-Table we determine the probability for P ( Z < .73 ):

P ( Z < 0.73 ) = 0.7673

P ( x_sample mean < 20948 ) = 0.7673

Answer 2

0.7673

Explanation:

in this question, we are asked to calculate a probability.

firstly. we start by calculating the z score, then onwards, we check the normal distribution table.

please check attachment for full and complete answer