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1. Find the mean of the following data:

| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|----------|------|-------|-------|-------|-------|
| Frequency| 20 | 24 | 40 | 36 | 20 |

2. The median of the following frequency distribution is 35. Find the value of [tex]x[/tex]. Also, find the modal class:

| Class Interval (CI) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---------------------|------|-------|-------|-------|-------|-------|-------|
| Frequency (f) | 2 | 3 | x | 6 | 5 | 3 | 2 |

3. If the mean of the four numbers 2, 6, 7, and [tex]a[/tex] is 15, and the mean of another set of five numbers 6, 18, 1, [tex]a[/tex], and [tex]b[/tex] is 50, what is the value of [tex]b[/tex]?

4. The mean of the following frequency distribution is 62.8. Find the missing frequency [tex]x[/tex]:

| Class Interval (CI) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
|---------------------|------|-------|-------|-------|--------|---------|
| Frequency (f) | 5 | 8 | x | 12 | 7 | 8 |

Answer :

Let's solve each part of the question step by step:


  1. Finding the Mean of the First Set of Data:

    To find the mean of grouped data, we use the formula:
    [tex]\text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i}[/tex]
    where [tex]f_i[/tex] is the frequency of each class and [tex]x_i[/tex] is the mid-point of the class intervals.


    • Calculate the mid-point ([tex]x_i[/tex]) for each class:


      • 0-10: [tex]x_i = \frac{0 + 10}{2} = 5[/tex]

      • 10-20: [tex]x_i = \frac{10 + 20}{2} = 15[/tex]

      • 20-30: [tex]x_i = \frac{20 + 30}{2} = 25[/tex]

      • 30-40: [tex]x_i = \frac{30 + 40}{2} = 35[/tex]

      • 40-50: [tex]x_i = \frac{40 + 50}{2} = 45[/tex]



    • Multiply the mid-point by the frequency for each class and sum them up:
      [
      \begin{align*}

      & (20 \times 5) + (24 \times 15) + (40 \times 25) + (36 \times 35) + (20 \times 45) \
      & = 100 + 360 + 1000 + 1260 + 900 \
      & = 3620

      \end{align*}
      ]


    • Sum of frequencies: [tex]20 + 24 + 40 + 36 + 20 = 140[/tex]


    • Therefore, the mean is:
      [tex]\text{Mean} = \frac{3620}{140} \approx 25.86[/tex]




  2. Finding the Median and Modal Class:

    Given that the median is 35, we need to find [tex]x[/tex] in the data below:




    CI
    0-10
    10-20
    20-30
    30-40
    40-50
    50-60
    60-70



    f
    2
    3
    x
    6
    5
    3
    2



    • The cumulative frequency just before the median class should be less than [tex]n/2 = 35/2 = 17.5[/tex].


    • Calculate cumulative frequencies:


      1. [tex]0-10:[/tex] 2

      2. [tex]10-20:[/tex] 2 + 3 = 5

      3. [tex]20-30:[/tex] 5 + [tex]x[/tex]

      4. [tex]30-40:[/tex] [tex]5 + x + 6 = 11 + x[/tex]




    Since the median is 35, the median class should be [tex]30-40[/tex], so:
    [tex]11 + x = 17.5[/tex]

    Solving this equation:
    [tex]x = 17.5 - 11 = 6.5[/tex]


    • Modal Class: The modal class is the class with the highest frequency. From the given data, frequencies are 2, 3, 6.5, 6, 5, 3, 2, so the modal class is [tex]30-40[/tex] with frequency 6.



  3. Finding the Value of [tex]b[/tex] Given Means of Two Groups:


    • First set: Mean of 2, 6, 7, and [tex]a[/tex] is 15:
      [tex]\frac{2 + 6 + 7 + a}{4} = 15[/tex]

      [tex]2 + 6 + 7 + a = 60 \Rightarrow a = 60 - 15 = 45[/tex]


    • Second set: Mean of 6, 18, 1, [tex]a[/tex], [tex]b[/tex] is 50:
      [tex]\frac{6 + 18 + 1 + a + b}{5} = 50[/tex]

      [tex]25 + 45 + b = 250 \Rightarrow b = 250 - 70 = 180[/tex]




  4. Finding the Missing Frequency [tex]x[/tex]:


    • Given mean is 62.8 for the data:





    CI
    0-20
    20-40
    40-60
    60-80
    80-100
    100-120



    f
    5
    8
    x
    12
    7
    8



    • Mid-point [tex]x_i[/tex] calculations:


      • 0-20: [tex]x_i = 10[/tex]

      • 20-40: [tex]x_i = 30[/tex]

      • 40-60: [tex]x_i = 50[/tex]

      • 60-80: [tex]x_i = 70[/tex]

      • 80-100: [tex]x_i = 90[/tex]

      • 100-120: [tex]x_i = 110[/tex]



    • Mean formula gives:
      [tex]\frac{(5 \times 10) + (8 \times 30) + (x \times 50) + (12 \times 70) + (7 \times 90) + (8 \times 110)}{5 + 8 + x + 12 + 7 + 8} = 62.8[/tex]
      The total sum of frequencies is [tex]40 + x[/tex].



    Calculate numerator:

    [tex]50 + 240 + 50x + 840 + 630 + 880 = 2640 + 50x[/tex]

    [tex]\frac{2640 + 50x}{40 + x} = 62.8[/tex]

    Solve for [tex]x[/tex]:

    [tex]2640 + 50x = 62.8(40 + x)[/tex]

    [tex]2640 + 50x = 2512 + 62.8x[/tex]

    [tex]128 = 12.8x[/tex]

    [tex]x = \frac{128}{12.8} = 10[/tex]



All calculations were performed to the best of understanding and aligning with the information provided in this context. Feel free to ask if you have further questions!

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