part two: How would I find... If only scores that are 1 standard deviation above the mean are considered for prizes, how

many prizes would be handed out?

2 months ago

Answered By Leonardo F

For part 1: in order to find the standard deviation of the math scores, we need to find the average or mean:

mean = (5+7+9+6+10+8+7+9+2+3+4+7+7+6)/14

mean = 6.43

Now, to find the standard deviation, we need to subtract each math score by the mean and square the resulting value, then divide by the number of scores, like this:

Part 2: we can determine the z score of each math score, like this:

z score = (value - mean)/std deviation

If the z score is higher than 1, then the value is to be considered for prizes. For example, let's condider the 9 score:

z score = (9 - 6.43)/2.23 = 1.15

So the 9 must be included.

If we do this for the entire set, only the values 9 and 10 are 1 std deviation above the mean. Since there are two 9's and one 10, we have to hand out 3 prizes in total.

2 months ago

## Answered By Leonardo F

For part 1: in order to find the standard deviation of the math scores, we need to find the average or mean:

mean = (5+7+9+6+10+8+7+9+2+3+4+7+7+6)/14

mean = 6.43

Now, to find the standard deviation, we need to subtract each math score by the mean and square the resulting value, then divide by the number of scores, like this:

Std deviation = SQRT{[ (5-6.43)^2+(7-6.43)^2+(9-6.43)^2+(6-6.43)^2+(10-6.43)^2+(8-6.43)^2+(7-6.43)^2+(9-6.43)^2+(2-6.43)^2+(3-6.43)^2+(4-6.43)^2+(7-6.43)^2+(7-6.43)^2+(6-6.43)^2 ] / 14}

SQRT means square root.

Std deviation = 2.23

Part 2: we can determine the z score of each math score, like this:

z score = (value - mean)/std deviation

If the z score is higher than 1, then the value is to be considered for prizes. For example, let's condider the 9 score:

z score = (9 - 6.43)/2.23 = 1.15

So the 9 must be included.

If we do this for the entire set, only the values 9 and 10 are 1 std deviation above the mean. Since there are two 9's and one 10, we have to hand out 3 prizes in total.