Question

the average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.5. assume that the sample is taken from a large population and the correction factor can be ignored. use la tabla de distribución normal estándar. round the final answers to four decimal places and intermediate z value calculations to two decimal places.
parte 1 de 3
find the probability that the mean of a sample of 25 people is less than 10:
p(\bar{x} < 10) = 0.0021
parte 2 de 3
find the probability that the mean of a sample of 25 people is greater than 10.
p(\bar{x} > 10) =

Explanation:

Step1: Identify parameters

Population mean $\mu=12$, population standard deviation $\sigma=3.5$, sample size $n=25$, sample mean $\bar{x}=10$.

Step2: Calculate standard error

Standard error $SE = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{25}} = 0.7$.

Step3: Compute z-score

$z = \frac{\bar{x} - \mu}{SE} = \frac{10 - 12}{0.7} \approx -2.86$.

Step4: Find $P(Z < -2.86)$

From standard normal table, $P(Z < -2.86) \approx 0.0021$.

Step5: Calculate $P(\bar{X} > 10)$

$P(\bar{X} > 10) = 1 - P(\bar{X} < 10) = 1 - 0.0021 = 0.9979$.

Answer:

0.0021
0.9979