Study the following data showing the age of the Primary School pupils at TumainiPrimary School and answer the questions that follow: 15, 8, 7, 6, 12, 5, 14 and 13.

(i) Determine the range and median of the age of the pupils.

(ii) Calculate the Standard Deviation.

ANSWER

(i) To determine the range and median of the age of the pupils:

First, let’s arrange the ages in ascending order: 5, 6, 7, 8, 12, 13, 14, 15

Range: The range is the difference between the highest and lowest values in the dataset. In this case, the highest value is 15, and the lowest value is 5. Range = 15 – 5 = 10

Median: The median is the middle value in a dataset when it is arranged in ascending or descending order. If there is an even number of values, the median is the average of the two middle values.

In this case, there are 8 ages. The middle two values are 8 and 12. Therefore, the median is the average of 8 and 12. Median = (8 + 12) / 2 = 10

(ii) To calculate the Standard deviation:

Step 1: Find the mean (average) of the ages. Mean = (5 + 6 + 7 + 8 + 12 + 13 + 14 + 15) / 8 = 10

Step 2: Calculate the deviations from the mean by subtracting the mean from each age value:

Deviations: -5, -4, -3, -2, 2, 3, 4, 5

Step 3: Square each deviation value:

Squared Deviations: 25, 16, 9, 4, 4, 9, 16, 25

Step 4: Find the mean of the squared deviations (variance): Variance = (25 + 16 + 9 + 4 + 4 + 9 + 16 + 25) / 8 = 14

Step 5: Take the square root of the variance to find the standard deviation:

Standard Deviation = √(Variance) = √14 ≈ 3.74

Therefore, the standard deviation of the ages of the pupils is approximately 3.74.

The post CSEE NECTA 2013 QUESTION 4 CALCULATION OF RANGE,MEDIAN AND STANDARD DEVIATION FOR UNGROUPED DATA appeared first on GEOGRAPHY POINT - YOUR GATEWAY TO GLOBAL GEOGRAPHY.