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Serga [27]
3 years ago
6

11

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0
((79-65)/65) x 100 = 21.54%
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The mean per capita income is 23,037 dollars per annum with a variance of 149,769. What is the probability that the sample mean
mihalych1998 [28]

Answer:

Probability that the sample mean would be less than 23,013 dollars is 0.2358.

Step-by-step explanation:

We are given that the mean per capita income is 23,037 dollars per annum with a variance of 149,769.

Also, a sample of 134 persons is randomly selected.

Firstly, Let \bar X = sample mean

The z-score probability distribution for sample mean is given by;

              Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean per capita income = 23,037 dollars p.a.

           \sigma = standard deviation = \sqrt{Variance} = \sqrt{149,769} = 387 dollars p.a

           n = sample of persons = 134

So, probability that the sample mean would be less than 23,013 dollars is given by = P(\bar X < 23,013 dollars)

      P(\bar X < 23,013) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{23,013-23,037}{\frac{387}{\sqrt{134} } } ) = P(Z < -0.72) = 1 - P(Z \leq 0.72)

                                                                    = 1 - 0.7642 = 0.2358

The above probability is calculated using the z-score table.

<em>Therefore, probability that the sample mean would be less than 23013 dollars if a sample of 134 persons is randomly selected is 0.2358.</em>

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