Question

5. The mean as a balance pointThe (incomplete) histogram for a sample consisting of n = 6 scores is shown on the following graph. The scores are represented as boxes sitting on a seesaw, but one of the values is missing from the graph. When all six values are sitting on the seesaw, it is balanced (rests level) when it pivots at a point equal to the sample mean M = 8.Missing ScoreM345678910Complete the following table for the five values that are displayed on the graph.ScoreDistance from the MeanX = 3point(s) the meanX = 6point(s) the meanX = 10point(s) the meanX = 10point(s) the meanX = 10point(s) the meanComplete the following statements for the five values that are displayed on the graph.The total distance below the mean ispoints.The total distance above the mean ispoints.Drag the orange box (square symbols) that represents the missing value onto the graph and correctly position it on the seesaw. (Hint: Use your answers and the fact that the mean balances the distances to determine the placement of the box.)

Answer 1

The missing value should be placed at point 9 which is 1 point above the mean.

Given that :

Sample mean = 8

Number of observations, n = 6

Scores and distances from the mean

Score (X) __ Distance from the mean

3 _________ 3 - 8 = - 5 points below the mean

6 _________ 6 - 8 = - 2 points below the mean

10 ________ 10 - 8 = 2 points above the mean

10 ________ 10 - 8 = 2 points above the mean

10 ________ 10 - 8 = 2 points above the mean

Total distance below the mean = (-5 + - 2) = - 7

Total distance above the mean = (2 + 2 + 2) = 6

Let, the missing value = p

Sample mean = ΣX / n

8 = (3 + 6 + 10 + 10 + 10 + p) / 6

8 × 6 = 39 + p

48 = 39 + p

p = 48 - 39

p = 9

The missing value, p should be at point 9. Which is a distance (9-8) = 1 above the mean.

This will make the distance below and above the mean equal.

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