We read your photo as:

Calculate the mean , median ,mode, standard deviation.

Mathematics

ScanSolve AI Answer

Step-by-step solution

14 steps

Here are the calculations for the mean, median, mode, and standard deviation from the given data.

First, let's set up the table with the necessary columns for calculations.
The class intervals are 30-34, 35-39, 40-44, 45-49, 50-54.
The class width (cc) is 5 (e.g., 34−30+1=534 - 30 + 1 = 5).
The assumed mean (AA) is given as 34.2.

Class IntervalF (Frequency)x (Midpoint)d=x−Ad = x - Au=x−Acu = \frac{x - A}{c}FuFuFu2Fu^2
30-34103232−34.2=−2.232 - 34.2 = -2.2−2.25=−0.44\frac{-2.2}{5} = -0.44−4.4-4.41.936
35-39123737−34.2=2.837 - 34.2 = 2.82.85=0.56\frac{2.8}{5} = 0.566.723.7632
40-4484242−34.2=7.842 - 34.2 = 7.87.85=1.56\frac{7.8}{5} = 1.5612.4819.4688
45-49154747−34.2=12.847 - 34.2 = 12.812.85=2.56\frac{12.8}{5} = 2.5638.498.304
50-5455252−34.2=17.852 - 34.2 = 17.817.85=3.56\frac{17.8}{5} = 3.5617.863.368
Total∑F=50\sum F = 50∑Fu=70.9\sum Fu = 70.9∑Fu2=186.84\sum Fu^2 = 186.84

1. Calculate Mean (xˉ\bar{x}):

  1. Use the formula for the assumed mean method.

    xˉ=A+(∑Fu∑F)×c\bar{x} = A + \left(\frac{\sum Fu}{\sum F}\right) \times c
  2. Substitute the values from the table.

    xˉ=34.2+(70.950)×5\bar{x} = 34.2 + \left(\frac{70.9}{50}\right) \times 5
  3. Simplify the expression.

    xˉ=34.2+70.910\bar{x} = 34.2 + \frac{70.9}{10} xˉ=34.2+7.09\bar{x} = 34.2 + 7.09 xˉ=41.29\bar{x} = 41.29

    The mean is 41.29\boxed{41.29}.

    2. Calculate Median:

  4. Find the total frequency (NN) and the median position.

    N=∑F=50N = \sum F = 50 Median position=N2=502=25th value\text{Median position} = \frac{N}{2} = \frac{50}{2} = 25^{th} \text{ value}
  5. Determine the median class using cumulative frequency (CF).

    Class IntervalFCF
    30-341010
    35-391222
    40-44830(The 25th value falls in this class)
    45-491545
    50-54550
    The median class is 40-44.
  6. Identify the values for the median formula.

    Lower boundary of median class (LL) = 39.5
    Frequency of median class (fmf_m) = 8
    Cumulative frequency of the class before the median class (CFbCF_b) = 22
    Class width (cc) = 5

  7. Apply the median formula for grouped data.

    Median=L+(N2−CFbfm)×c\text{Median} = L + \left(\frac{\frac{N}{2} - CF_b}{f_m}\right) \times c Median=39.5+(25−228)×5\text{Median} = 39.5 + \left(\frac{25 - 22}{8}\right) \times 5 Median=39.5+(38)×5\text{Median} = 39.5 + \left(\frac{3}{8}\right) \times 5 Median=39.5+158\text{Median} = 39.5 + \frac{15}{8} Median=39.5+1.875\text{Median} = 39.5 + 1.875 Median=41.375\text{Median} = 41.375

    The median is 41.375\boxed{41.375}.

    3. Calculate Mode:

  8. Identify the modal class (class with the highest frequency).

    The highest frequency is 15, which corresponds to the class 45-49.
    Modal class = 45-49.

  9. Identify the values for the mode formula.

    Lower boundary of modal class (LL) = 44.5
    Frequency of modal class (fmf_m) = 15
    Frequency of the class before the modal class (f1f_1) = 8 (frequency of 40-44)
    Frequency of the class after the modal class (f2f_2) = 5 (frequency of 50-54)
    Class width (cc) = 5

  10. Apply the mode formula for grouped data.

    Mode=L+(fm−f1(fm−f1)+(fm−f2))×c\text{Mode} = L + \left(\frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)}\right) \times c Mode=44.5+(15−8(15−8)+(15−5))×5\text{Mode} = 44.5 + \left(\frac{15 - 8}{(15 - 8) + (15 - 5)}\right) \times 5 Mode=44.5+(77+10)×5\text{Mode} = 44.5 + \left(\frac{7}{7 + 10}\right) \times 5 Mode=44.5+(717)×5\text{Mode} = 44.5 + \left(\frac{7}{17}\right) \times 5 Mode=44.5+3517\text{Mode} = 44.5 + \frac{35}{17} Mode=44.5+2.05882...\text{Mode} = 44.5 + 2.05882... Mode≈46.56\text{Mode} \approx 46.56

    The mode is 46.56\boxed{46.56}.

    4. Calculate Standard Deviation (σ\sigma):

  11. Use the formula for standard deviation with the step deviation method.

    σ=c×∑Fu2∑F−(∑Fu∑F)2\sigma = c \times \sqrt{\frac{\sum Fu^2}{\sum F} - \left(\frac{\sum Fu}{\sum F}\right)^2}
  12. Substitute the values from the table.

    σ=5×186.8450−(70.950)2\sigma = 5 \times \sqrt{\frac{186.84}{50} - \left(\frac{70.9}{50}\right)^2}
  13. Perform the calculations inside the square root.

    σ=5×3.7368−(1.418)2\sigma = 5 \times \sqrt{3.7368 - (1.418)^2} σ=5×3.7368−2.010724\sigma = 5 \times \sqrt{3.7368 - 2.010724} σ=5×1.726076\sigma = 5 \times \sqrt{1.726076}
  14. Calculate the square root and multiply by cc.

    σ=5×1.313794...\sigma = 5 \times 1.313794... σ≈6.57\sigma \approx 6.57

    The standard deviation is 6.57\boxed{6.57}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question