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the average number of moves a person makes in his or her lifetime is 12…

Question

the average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.6. assume that the sample is taken from a large population and the correction factor can be ignored. use the standard normal distribution table. round the final answers to four decimal places and intermediate z value calculations to two decimal places.
part 1 of 3
find the probability that the mean of a sample of 25 people is less than 10:
$p(\overline{x} < 10) = 0.0052$
part 2 of 3
find the probability that the mean of a sample of 25 people is greater than 10.
$p(\overline{x} > 10) = 0.9948$

Explanation:

Step1: Calculate standard error

Standard error $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{25}} = 0.72$

Step2: Compute z-score for $\bar{X}=10$

$z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{10 - 12}{0.72} \approx -2.78$

Step3: Find $P(Z < -2.78)$

From standard normal table, $P(Z < -2.78) = 0.0027$

Step4: Find $P(\bar{X} > 10)$

$P(\bar{X} > 10) = 1 - P(\bar{X} < 10) = 1 - 0.0027 = 0.9973$

Answer:

Part 1: 0.0027
Part 2: 0.9973