Calculate Standard Deviation B.Ed Notes

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Calculate Standard Deviation

Standard Deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the value tends to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

Question 1: Calculate the Standard Deviation of the following scores: 52, 50, 56, 68, 65, 62, 57, 70

Solution:

Mean = (Sum of all the observations)/Total number of observations

Mean = 482/8 = 60

Calculate Standard Deviation (Assignment)

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Calculate Standard Deviation B.Ed Notes

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Calculate Standard Deviation B.Ed Notes

Conclusion

  • The standard deviation can be denoted as σ or Sigma.
  • Standard deviation is regarded as the most stable and reliable measure of variability as it employs a mean for its computation.
  • Standard deviation gives the standard answers and avoids fluctuations or deviations.

Question 2: A supermarket wants to know the average amount of money spent by their customers per purchase. They randomly sampled 10 customers and recorded the amount of money they spent in dollars: ₹20, ₹15, ₹25, ₹30, ₹18, ₹22, ₹28, ₹19, ₹27, ₹23. Calculate the standard deviation for the data.

Solution:

To calculate the standard deviation, we first need to calculate the mean:

Mean = (20 + 15 + 25 + 30 + 18 + 22 + 28 + 19 + 27 + 23) / 10 = 23

Next, we need to calculate the variance, which is the sum of the squared differences between each data point and the mean divided by the number of data points:

Variance = ((20 – 23)^2 + (15 – 23)^2 + (25 – 23)^2 + (30 – 23)^2 + (18 – 23)^2 + (22 – 23)^2 + (28 – 23)^2 + (19 – 23)^2 + (27 – 23)^2 + (23 – 23)^2) / 10

Variance = (9 + 64 + 4 + 49 + 25 + 1 + 25 + 16 + 16 + 0) / 10 = 20.9

Finally, we can calculate the standard deviation by taking the square root of the variance:

Standard Deviation = sqrt(20.9) = 4.57

Therefore, the standard deviation for the amount of money spent by the supermarket’s customers is ₹4.57.


Question 3: A teacher recorded the test scores of 8 students in a class. The scores are as follows: 80, 85, 90, 95, 75, 70, 88, 92. Calculate the standard deviation of the test scores.

Solution: First, calculate the mean:

Mean = (80 + 85 + 90 + 95 + 75 + 70 + 88 + 92) / 8 = 85

Next, calculate the variance:

Variance = ((80 – 85)^2 + (85 – 85)^2 + (90 – 85)^2 + (95 – 85)^2 + (75 – 85)^2 + (70 – 85)^2 + (88 – 85)^2 + (92 – 85)^2) / 8

Variance = (25 + 0 + 25 + 100 + 100 + 225 + 9 + 49) / 8 = 68.25

Finally, calculate the standard deviation:

Standard Deviation = sqrt(68.25) = 8.26

Therefore, the standard deviation of the test scores is 8.26.


Q 4. A company wants to know the average daily sales of their store. They randomly sampled the sales of 12 days and recorded the amount in dollars: 1500, 1700, 1800, 1900, 1600, 1750, 1650, 1850, 1950, 2000, 1550, 1680. Calculate the standard deviation of the daily sales.

Solution:

First, calculate the mean:

Mean = (1500 + 1700 + 1800 + 1900 + 1600 + 1750 + 1650 + 1850 + 1950 + 2000 + 1550 + 1680) / 12 = 1750

Next, calculate the variance:

Variance = ((1500 – 1750)^2 + (1700 – 1750)^2 + (1800 – 1750)^2 + (1900 – 1750)^2 + (1600 – 1750)^2 + (1750 – 1750)^2 + (1650 – 1750)^2 + (1850 – 1750)^2 + (1950 – 1750)^2 + (2000 – 1750)^2 + (1550 – 1750)^2 + (1680 – 1750)^2) / 12

Variance = (25000 + 10000 + 2500 + 22500 + 22500 + 0 + 10000 + 10000 + 40000 + 62500 + 40000 + 4900) / 12 = 18208.33

Finally, calculate the standard deviation: Standard Deviation = sqrt(18208.33) = 134.85 Therefore, the standard deviation of the daily sales is 134.85


Q 5. A car rental company wants to know the average age of their customers. They randomly sampled the age of 15 customers and recorded the results: 25, 28, 35, 30, 42, 23, 29, 27, 31, 36, 24, 26, 34, 33, 37. Calculate the standard deviation of the customer ages.

Solution:

First, calculate the mean:

Mean = (25 + 28 + 35 + 30 + 42 + 23 + 29 + 27 + 31 + 36 + 24 + 26 + 34 + 33 + 37) / 15

Mean = 30

Next, calculate the variance:

Variance = ((25 – 30)^2 + (28 – 30)^2 + (35 – 30)^2 + (30 – 30)^2 + (42 – 30)^2 + (23 – 30)^2 + (29 – 30)^2 + (27 – 30)^2 + (31 – 30)^2 + (36 – 30)^2 + (24 – 30)^2 + (26 – 30)^2 + (34 – 30)^2 + (33 – 30)^2 + (37 – 30)^2) / 15

Variance = (25 + 4 + 25 + 0 + 144 + 49 + 1 + 9 + 1 + 36 + 36 + 16 + 16 + 9 + 49) / 15

The Variance: 28.4

Finally, calculate the standard deviation:

Standard Deviation = sqrt(28.4) = 5.33

Therefore, the standard deviation of the customer ages is 5.33 years.

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