Question

If Z follows a standard normal distribution, determine the cutoff score if P( Z < a ) = 0.575. Draw a density curve that represents this information and use the calculator to determine the answer. Write the calculator function

Graduate Management Aptitude Test (GMAT) scores are widely used by graduate schools of business as an entrance requirement. Let the random variable X represent the score on the GMAT and follow a normal distribution with a mean score of 548, standard deviation of 105. What is the probability of getting a score of more than 500? Use the calculator to determine the probability

Graduate Management Aptitude Test (GMAT) scores are widely used by graduate schools of business as an entrance requirement. Let the random variable X represent the score on the GMAT and follow a normal distribution with a mean score of 548, standard deviation of 105. It is reported that the minimum score for admittance to a prestigious business school is scoring in the 96th Percentile. Determine the minimum score on the GMAT of someone who was admitted to this school. Use the calculator to determine the answer. Write the calculator function

Answer 1

a) a=0.181912

b) The probability of getting a score of more than 500 is P=0.677

c) Minimum score to be admitted = 732

Explanation:

a) The cutoff value a can be looked-up from a standard normal distribution table or applet.

The value for a to have P(Z<a)=0.575 is a=0.18912.

b) To calculate the probability of a value in a not-standard normal distribution, we can transform it to the standard form and look up for the value of the z-score.

If the distribution has a mean of 548 and a standard deviation of 105, the probabilty o getting a score of more than 500 is:

`z=(x-\mu)/\sigma=(500-548)/105=-48/105=-0.46\\\\P(x>500)=P(z>-0.46)=0.677`

The probability of getting a score of more than 500 is P=0.677.

c) Scoring more than the 96th percentile means that only will be admitted if the score is higher than the 96% of the population of scores.

As it is a normal distribution, we first calculate the z-score for which 96% of the values lies below:

`P(z<z*)=0.96\\\\z*=1.75`

Then, we convert this z-score to the score distribution as:

`X=\mu+z*\sigma=548+1.75*105=548+183.75=731.75\approx 732`