Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 61 graduates. Find the probability that a single randomly selected graduate has a salary between 158203.2 and 164028.9 dollars. P(158203.2 < X < 164028.9) = (Enter your answers as numbers accurate to 4 decimal places.)

Assignment Question

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 147000 dollars. Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 61 graduates. Find the probability that a single randomly selected graduate has a salary between 158203.2 and 164028.9 dollars. P(158203.2 < X < 164028.9) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n=61 has a mean value between 158203.2 and 164028.9 dollars. P(158203.2 < M < 164028.9) = (Enter your answers as numbers accurate to 4 decimal places.)

Assignment Answer

To find the probability that a single randomly selected graduate has a salary between 158203.2 and 164028.9 dollars, you can use the standard normal distribution. First, you need to standardize the values using the formula:

�=�−��Z=σX−μ​

where:

• $X$ is the individual salary,
• $\mu$ is the mean annual salary (147000 dollars),
• $\sigma$ is the standard deviation (35000 dollars).

So, for the lower value (158203.2):

�1=158203.2−14700035000Z1​=35000158203.2−147000​

And for the upper value (164028.9):

�2=164028.9−14700035000Z2​=35000164028.9−147000​

Once you have �1Z1​ and �2Z2​, you can look up these values in a standard normal distribution table or use a calculator to find the probabilities.

For the probability that a single randomly selected graduate has a salary between 158203.2 and 164028.9 dollars:

�(158203.2<�<164028.9)=�(�1<�<�2)P(158203.2<X<164028.9)=P(Z1​<Z<Z2​)

Now, for the probability that a random sample of size $n=61$ has a mean value between 158203.2 and 164028.9 dollars, you need to use the standard error of the mean ($SE$) formula:

��=��SE=n​σ​

Substitute the given values:

��=3500061SE=61​35000​

Now, you can standardize the values using the standard normal distribution:

�1′=158203.2−���Z1′​=SE158203.2−μ​ �2′=164028.9−���Z2′​=SE164028.9−μ​

And the probability is given by:

�(158203.2<�<164028.9)=�(�1′<�<�2′)P(158203.2<M<164028.9)=P(Z1′​<Z<Z2′​)

Now, calculate these values and probabilities, and enter your answers as numbers accurate to 4 decimal places.

Frequently Asked Questions

Q: How is the probability calculated for a single graduate’s salary range?

A: The probability is calculated using the standard normal distribution by standardizing the values of the salary range and finding the probability between these standardized values.

Q: What are the key parameters for calculating the probability of a single graduate’s salary?

A: The mean annual salary ($\mu$) of graduates after 10 years, the standard deviation ($\sigma$) of salaries, and the specific salary range are crucial parameters.

Q: How is the standard error of the mean calculated for a sample of size 61?

A: The standard error ($SE$) is calculated as the standard deviation divided by the square root of the sample size (�n​).

Q: What does it mean to standardize values in statistical analysis?

A: Standardizing involves converting raw data into z-scores, which represent the number of standard deviations a data point is from the mean.

Q: Why is the category “Education and Career Statistics” suitable for this analysis?

A: This category encompasses the statistical analysis of salary outcomes for MBA graduates, linking education (MBA programs) and career (salary distributions) aspects.