High School

Find the standard deviation of the following distribution:

| Age | No. of Persons |
|-------|----------------|
| 20-25 | 170 |
| 25-30 | 110 |
| 30-35 | 80 |
| 35-40 | 45 |
| 40-45 | 40 |
| 45-50 | 35 |

Assume the average is 32.5.

Answer :

To find the standard deviation of the given frequency distribution, follow these steps:

  1. Identify the Class Midpoints: The midpoint of each class interval is calculated by taking the average of the lower and upper limits of each class.

    • For 20-25: Midpoint is [tex]\frac{20 + 25}{2} = 22.5[/tex]
    • For 25-30: Midpoint is [tex]\frac{25 + 30}{2} = 27.5[/tex]
    • For 30-35: Midpoint is [tex]\frac{30 + 35}{2} = 32.5[/tex]
    • For 35-40: Midpoint is [tex]\frac{35 + 40}{2} = 37.5[/tex]
    • For 40-45: Midpoint is [tex]\frac{40 + 45}{2} = 42.5[/tex]
    • For 45-50: Midpoint is [tex]\frac{45 + 50}{2} = 47.5[/tex]
  2. Calculate Deviations: Subtract the assumed average (32.5) from each class midpoint.

  3. Deviation from Assumed Average:

    • For 20-25: [tex]22.5 - 32.5 = -10[/tex]
    • For 25-30: [tex]27.5 - 32.5 = -5[/tex]
    • For 30-35: [tex]32.5 - 32.5 = 0[/tex]
    • For 35-40: [tex]37.5 - 32.5 = 5[/tex]
    • For 40-45: [tex]42.5 - 32.5 = 10[/tex]
    • For 45-50: [tex]47.5 - 32.5 = 15[/tex]
  4. Calculate Squared Deviations and multiply by the frequency of each class:

    • For 20-25: [tex](-10)^2 \times 170 = 17000[/tex]
    • For 25-30: [tex](-5)^2 \times 110 = 2750[/tex]
    • For 30-35: [tex](0)^2 \times 80 = 0[/tex]
    • For 35-40: [tex]5^2 \times 45 = 1125[/tex]
    • For 40-45: [tex]10^2 \times 40 = 4000[/tex]
    • For 45-50: [tex]15^2 \times 35 = 7875[/tex]
  5. Sum Up the Squared Deviations:

    • Total = [tex]17000 + 2750 + 0 + 1125 + 4000 + 7875 = 32750[/tex]
  6. Find the Total Number of Persons (N):

    • [tex]170 + 110 + 80 + 45 + 40 + 35 = 480[/tex]
  7. Calculate Variance:
    [tex]\text{Variance} = \frac{32750}{480} \approx 68.23[/tex]

  8. Calculate Standard Deviation:
    [tex]\text{Standard Deviation} = \sqrt{68.23} \approx 8.26[/tex]

So, the standard deviation of the distribution is approximately 8.26 years.