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Calculate the standard deviation of the given data using the assumed mean method and state the frequency of the modal class.

Mathematics

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Answers

4 parts
a

15\text{15}

b(i)

17.434 cm\text{17.434 cm}

b(ii)

21.755 cm2\text{21.755 cm}^2

b(iii)

4.664 cm\text{4.664 cm}

Step-by-step solution

8 steps

To solve this problem, we first need to organize the given data and calculate the midpoints of each class interval. We will also calculate the products fdfd and fd2fd^2 for the assumed mean method.

Given data:

Length (cm)No. of rods (f)
8-104
11-137
14-1611
17-1915
20-228
23-255
26-283

Let A=18A = 18 be the assumed mean.
We calculate the midpoint (xx) for each class, the deviation d=x−Ad = x - A, fdfd, and fd2fd^2.

Length (cm)ffMidpoint (xx)d=x−18d = x - 18fdfdfd2fd^2
8-1049-9-36324
11-13712-6-42252
14-161115-3-3399
17-191518000
20-2282132472
23-25524630180
26-28327927243
Total∑f=53\sum f = 53∑fd=−30\sum fd = -30∑fd2=1170\sum fd^2 = 1170

Part a· 2 steps

State the frequency of the modal class.

  1. Identify the modal class.

    The modal class is the class interval with the highest frequency. From the table, the highest frequency is 15, which corresponds to the class interval 17-19.

  2. State the frequency.

    The frequency of the modal class is 15.
    The frequency of the modal class is 15\boxed{\text{15}}.

Part b(i)

17.434 cm\text{17.434 cm}

Actual mean

Using 18 as an assumed mean, calculate

  1. Use the formula for the actual mean with an assumed mean.

    The formula is xˉ=A+∑fd∑f\bar{x} = A + \frac{\sum fd}{\sum f}.

  2. Substitute the values.

    xˉ=18+−3053\bar{x} = 18 + \frac{-30}{53} xˉ=18−0.5660377...\bar{x} = 18 - 0.5660377... xˉ≈17.434\bar{x} \approx 17.434

    The actual mean is 17.434 cm\boxed{\text{17.434 cm}}.

Part b(ii)

21.755 cm2\text{21.755 cm}^2

Variance

  1. Use the formula for variance with an assumed mean.

    The formula is σ2=∑fd2∑f−(∑fd∑f)2\sigma^2 = \frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2.

  2. Substitute the values.

    σ2=117053−(−3053)2\sigma^2 = \frac{1170}{53} - \left(\frac{-30}{53}\right)^2 σ2=22.075471698−(−0.5660377358)2\sigma^2 = 22.075471698 - (-0.5660377358)^2 σ2=22.075471698−0.320409009\sigma^2 = 22.075471698 - 0.320409009 σ2=21.755062689\sigma^2 = 21.755062689 σ2≈21.755\sigma^2 \approx 21.755

    The variance is 21.755 cm2\boxed{\text{21.755 cm}^2}.

Part b(iii)

4.664 cm\text{4.664 cm}

Standard deviation

  1. Use the formula for standard deviation.

    The formula is σ=σ2\sigma = \sqrt{\sigma^2}.

  2. Substitute the variance value.

    σ=21.755062689\sigma = \sqrt{21.755062689} σ=4.6642319...\sigma = 4.6642319... σ≈4.664\sigma \approx 4.664

    The standard deviation is 4.664 cm\boxed{\text{4.664 cm}}.

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