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skelet666 [1.2K]
3 years ago
7

The mean per capita income is 20,908 dollars per annum with a standard deviation of 407 dollars per annum. What is the probabili

ty 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.

Mathematics
2 answers:
Maurinko [17]3 years ago
8 0

Answer:

0.7673

Step-by-step 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

larisa86 [58]3 years ago
4 0

Answer:

P  (  x_sample mean  <  20948 ) = 0.7673

Step-by-step 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

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