Question 1168176: Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 159000 dollars. Assume the standard deviation is 40000 dollars. Suppose you take a simple random sample of 97 graduates.
Find the probability that a single randomly selected policy has a mean value between 146815.8 and 156969.3 dollars.
P(146815.8 < X < 156969.3) =
Find the probability that a random sample of size
n=97 has a mean value between 146815.8 and 156969.3 dollars.
P(146815.8 < M < 156969.3) =
(Enter your answers as numbers accurate to 4 decimal places.)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $\mu$ be the mean annual salary for graduates 10 years after graduation, so $\mu = 159000$.
Let $\sigma$ be the standard deviation of the annual salary, so $\sigma = 40000$.
**Find the probability that a single randomly selected graduate has a mean value between $146815.8 and $156969.3 dollars.**
For a single randomly selected graduate (n=1), the distribution is normal with mean $\mu = 159000$ and standard deviation $\sigma = 40000$. We want to find $P(146815.8 < X < 156969.3)$.
First, we calculate the z-scores for the lower and upper bounds:
$z_1 = \frac{146815.8 - 159000}{40000} = \frac{-12184.2}{40000} \approx -0.3046$
$z_2 = \frac{156969.3 - 159000}{40000} = \frac{-2030.7}{40000} \approx -0.0508$
Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or a calculator:
$P(Z < -0.3046) \approx 0.3803$
$P(Z < -0.0508) \approx 0.4797$
The probability $P(146815.8 < X < 156969.3) = P(-0.3046 < Z < -0.0508) = P(Z < -0.0508) - P(Z < -0.3046) = 0.4797 - 0.3803 = 0.0994$.
$P(146815.8 < X < 156969.3) = \boxed{0.0994}$
**Find the probability that a random sample of size n=97 has a mean value between $146815.8 and $156969.3 dollars.**
For a sample of size $n=97$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}} = \mu = 159000$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{40000}{\sqrt{97}} \approx \frac{40000}{9.84885} \approx 4061.39$.
Now, we calculate the z-scores for the lower and upper bounds of the sample mean:
$z_1 = \frac{146815.8 - 159000}{4061.39} = \frac{-12184.2}{4061.39} \approx -3.0000$
$z_2 = \frac{156969.3 - 159000}{4061.39} = \frac{-2030.7}{4061.39} \approx -0.5000$
Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or a calculator:
$P(Z < -3.0000) \approx 0.0013$
$P(Z < -0.5000) \approx 0.3085$
The probability $P(146815.8 < \bar{X} < 156969.3) = P(-3.0000 < Z < -0.5000) = P(Z < -0.5000) - P(Z < -3.0000) = 0.3085 - 0.0013 = 0.3072$.
$P(146815.8 < M < 156969.3) = \boxed{0.3072}$
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