## Question 484:

1## Answer:

No answer provided yet.You'll want to calculate a z-score for the scores provided. To calculate a z-score you subtract the data-point from the mean and divide that result by the standard deviation.

- For the 40th percentile, we first need to find the z-score for this percentile, then match the score given the mean and standard deviation. We'll assume they are talking about one-sided area. You can find the z-score in a z-table or use the percentile to z-score calculator and select 1-sided. We get a corresponding z-score of -0.2529. Now we setup an equation to solve for the unknown score x.
- (x-50)/10 = -0.2529
- x-50 = -2.529
- x = 47.471, so a score of 47.471 will be in the 40th percentile of scores.

- To find the scores above 70 we follow a similar procedure.
- First find the z-score for a score of 70
- (70-50)/10 = 2
- So a score of 70 is 2 standard deviations above the mean (which is what a z-score of 2 means). Now lookup the 1-sided area above 2 standard deviations using the z-score to percentile calculator. You should get 2.28%, or 2.28 percent of the scores would be above 70

- For the middle 50% we're interested in the 2-sided area. We want the 2-sided z-score for 50%. Entering .50, 2-sided in the percentile to z-score calculator we get the z-score of .6742, but we also want the opposite z-score of-.6472 since we wan the middle 50%, not 50% above or below this value.
- Setting up the equation we have
- (x-50)/10 = .6742
- x-50 = 6.742
- x = 56.742
- Since we want the middle 50% we also want the negative z-score of -.6742.
- (x-50)/10 -.6742
- x-50 = -6.742
- x = 43.258

So the middle 50 percent of scores would fall between around 43.3 and 56.74. It's easy to get lost in the z-score calculations (1-sided, 2-sided etc) and we can also visually verify this result by using the interactive graph of the standard normal curve. By entering the mean of 50 and standard deviation of 10 in the 1st graph and hovering the mouse over -.67 and .67 we see a shaded area of 50% and the scores of 43 and 56 confirming our computations.