Question 1168175: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 30 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 30 weeks and that the population standard deviation is 2.4 weeks. Suppose you would like to select a random sample of 59 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 29.7.
P(X > 29.7) =
(Enter your answers as numbers accurate to 4 decimal places.)
Find the probability that a sample of size
n=59 is randomly selected with a mean greater than 29.7.
P(M > 29.7) =
(Enter your answers as numbers accurate to 4 decimal places.)
I tried to complete the question and got these answers but they were wrong.
a) 0.4483
b) 0.1685
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $\mu$ be the population mean length of unemployment, so $\mu = 30$ weeks.
Let $\sigma$ be the population standard deviation, so $\sigma = 2.4$ weeks.
**Find the probability that a single randomly selected value is greater than 29.7.**
For a single randomly selected individual ($n=1$), the distribution is normal with mean $\mu = 30$ and standard deviation $\sigma = 2.4$. We want to find $P(X > 29.7)$.
First, we calculate the z-score for $x = 29.7$:
$z = \frac{x - \mu}{\sigma} = \frac{29.7 - 30}{2.4} = \frac{-0.3}{2.4} = -0.125$
Now, we find the probability $P(Z > -0.125)$, where $Z$ is a standard normal random variable.
$P(Z > -0.125) = 1 - P(Z \le -0.125)$
Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z \le -0.125) \approx 0.4502$
So, $P(Z > -0.125) = 1 - 0.4502 = 0.5498$.
$P(X > 29.7) = \boxed{0.5498}$
**Find the probability that a sample of size n=59 is randomly selected with a mean greater than 29.7.**
For a sample of size $n=59$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}} = \mu = 30$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{59}} \approx \frac{2.4}{7.6811} \approx 0.31245$.
Now, we calculate the z-score for the sample mean $\bar{X} = 29.7$:
$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{29.7 - 30}{0.31245} = \frac{-0.3}{0.31245} \approx -0.9602$
Now, we find the probability $P(Z > -0.9602)$, where $Z$ is a standard normal random variable.
$P(Z > -0.9602) = 1 - P(Z \le -0.9602)$
Looking up the value in a standard normal distribution table or using a calculator, we find:
$P(Z \le -0.9602) \approx 0.1685$
So, $P(Z > -0.9602) = 1 - 0.1685 = 0.8315$.
$P(M > 29.7) = \boxed{0.8315}$
Based on your previous answers:
a) 0.4483 is close to $P(Z \le -0.125) \approx 0.4502$, which is not the required probability for the first part.
b) 0.1685 is close to $P(Z \le -0.9602) \approx 0.1685$, which is $1 - P(M > 29.7)$.
Your calculations likely had a sign error or you were looking for the complementary probability.
Final Answer: The final answer is $\boxed{P(X > 29.7) = 0.5498, P(M > 29.7) = 0.8315}$
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