Question

For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 505 and a standard deviation of 170. The grading process of this year's exam has just begun. The average score of the 35 exams graded so far is 530.What is the probability that a sample of 35 exams will have a mean score of 530 or more if the exam scores follow the same distribution as in the past?

Answer 1

`P(\bar X>530)=1-0.808=0.192`

Explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:

`X \sim N(\mu=505,\sigma=170)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(505,(170)/(√(35)))`

2) Calculate the probability

We want this probability:

`P(\bar X>530)=1-P(\bar X<530)`

The best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/((\sigma)/(√(n)))`

If we apply this formula to our probability we got this:

`P(\bar X >530)=1-P(Z<(530-505)/((170)/(√(35))))=1-P(Z<0.87)`

`P(\bar X>530)=1-0.808=0.192`