Mathematics, 20.03.2020 03:02, uhhgray

Given that the ages (in years) of the residents in a city are normally distributed with a population mean of 37 and population standard deviation 13.
You randomly select 25 people. Follow the following steps to find the probability that their mean age is between 30 and 44.

(1)What is the distribution of sample mean? And why?
(2) What the mean and standard deviation for this sample mean?
(3) Find z score
(4) Find probability

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answered: Mune6745

(1) The distribution of the sample mean is Normal distribution

(2) Mean and Standard deviation for this sample mean is 37 and 2.6 respectively.

(3) z score is 2.69 and -2.69.

(4) Probability that their mean age is between 30 and 44 = 0.9928

Step-by-step explanation:

We are given that the ages (in years) of the residents in a city are normally distributed with a population mean of 37 and population standard deviation 13.

Also, 25 people are randomly selected.

(a) The distribution of the sample mean is Normal distribution because the sample of 25 people is taken only from the population data of the residents in a city which already follow normal distribution.

(b) Mean of the sample, $\bar X$ = population mean $\mu$ = 37

According Central limit theorem, Standard deviation of sample, s = population standard deviation ÷ $\sqrt{n}$

= $\frac{13}{\sqrt{25} }$ = 2.6

So, the z score probability distribution for sample mean is given by;

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

P(30 < $\bar X$ < 44) = P( $\bar X$ < 44) - P( $\bar X$ $\leq$ 30)

P( $\bar X$ < 44) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{44-37}{\frac{13}{\sqrt{25} } }$ ) = P(Z < 2.69) = 0.9964

P( $\bar X$ $\leq$ 30) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{30-37}{\frac{13}{\sqrt{25} } }$ ) = P(Z $\leq$ -2.69) = 1 - P(Z < 2.69)

= 1 - 0.9964 = 0.0036

Therefore, P(30 < $\bar X$ < 44) = 0.9964 - 0.0036 = 0.9928

answered: Guest

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answered: Guest

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