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Statistics Test - 13

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Statistics Test - 13
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Weekly Quiz Competition
  • Question 1
    1 / -0

    If  ∑fi = 15, ∑fx= 3p+36 and the mean of the distribution is 3, then the value of ‘p’ is

    Solution
    Mean \(=\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
    \(\Rightarrow 3=\frac{3 p+36}{15}\)
    \(\Rightarrow 3 p+36=45\)
    \(\Rightarrow 3 p=45-36\)
    \(\Rightarrow 3 p=9\)
    \(\Rightarrow p=\frac{9} {3}\)
    \(\Rightarrow p=3\)
  • Question 2
    1 / -0

    If xi′s are the midpoints of the class intervals of grouped data, fi′s are the corresponding frequencies and \(\bar{x}\) is the mean, then ∑(fixi − \(\bar{x}\))  is equal to

    Solution
    Given that, \(x_{i}\) are the mid points of the class intervals.
    \(f _{i}\) are the corresponding frequencies.
    The mean \(\bar{x}\) for the grouped data is \(\bar{x}=\frac{\sum f_{i} x_{i}}{n}\)
    \(\Longrightarrow \overline{ x } n =\sum f _{ i } x _{ i }\).........(1)
    \(\sum\left(f_{i} x_{i}-\bar{x}\right)=\sum f_{i} x_{i}-\sum \bar{x}\)
    \(\Longrightarrow \sum\left(f_{i} x_{i}-\bar{x}\right)=\bar{x} n-\sum \bar{x}(f r o m(1))\)
    \(\Longrightarrow \sum\left(f_{i} x_{i}-\bar{x}\right)=\bar{x} n-\bar{x} n=0\)
  • Question 3
    1 / -0

    The marks obtained by 9 students in Mathematics are 59, 46, 31, 23, 27, 40, 52, 35 and 29. The mean of the data is

    Solution
    Given: 59,46,31,23,27,40,52,35 and 29
    Mean \(=\frac{\text { Sum of all observations }}{\text { Number of oservation }}\)
    \(=\frac{59+46+31+23+27+40+52+35+29}{9}\)
    \(=\frac{342}{9}\)
    \(=38\)
  • Question 4
    1 / -0

    The mean of the data when ∑ fidi= −500, ∑ fi = 100 and a = 125 is

    Solution
    Mean \(=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}\)
    \(=125+\frac{-500}{100}=125-5=120\)
  • Question 5
    1 / -0

    If ∑ fiu= 29, ∑ f= 30, a = 47.5 and h = 15, then the value of x̄ is


    Solution
    \(\bar{x}=a+\frac{\sum f_{i} u_{i}}{\sum_{i}} \times h\)
    \(=47.5+\frac{29}{30} \times 15\)
    \(=47.5+\frac{29} {2}\)
    \(=47.5+14.5\)
    \(=62\)
  • Question 6
    1 / -0

    The mean of the first 10 prime numbers is

    Solution
    The first 10 prime numbers are 2,3,5,7,11,13,17 19,23,29
    \(\therefore\) Mean \(=\frac{\text { Sum of first } 10 \text { prime numbers }}{10}\) \(=\frac{2+3+5+7+11+13+17+19+23+29}{10}\)
    \(=\frac{129}{10}\)
    \(=12.9\)
  • Question 7
    1 / -0

    If ∑ fi=17, ∑ fixi=4p + 63 and the mean of the distribution is 7, then the value of ‘p’ is


    Solution
    since, Mean \(=\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
    \(\Rightarrow 7=\frac{4 p+63}{17}\)
    \(\Rightarrow 4 p+63=119\)
    \(\Rightarrow 4 p=119-63\)
    \(\Rightarrow 4 p=56\)
    \(\Rightarrow p=14\)
  • Question 8
    1 / -0

    ∑ (xi − x̄) is equal to

    Solution
    \(\sum x_{i}-\bar{x}=\sum x_{i}-\sum(\bar{x})\)
    \(\sum(\bar{x})= n (\bar{x})\) by definition
    But \(\bar{x}=\left(\sum x_{i}\right) / n\) by definition
    so \(\sum x_{i}-\sum(\bar{x})=\sum x_{i}- n \left(\left(\sum x_{i}\right) / n\right)\)
    which is equal to \(=\left(\sum x_{i}\right)-\left(\sum x_{i}\right)=0\)
  • Question 9
    1 / -0

    If ∑fixi = 1860 and ∑fi = 30, then the value of \( \bar{x}\) is

    Solution
    \(\bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
    \(=\frac{1860}{30}\)
    \(=62\)
  • Question 10
    1 / -0

    The mean of the first 10 natural numbers is


    Solution
    The first 10 natural numbers are \(1,2,3, \ldots \ldots \ldots, 10\) \(\therefore\)
    Mean \(=\frac{\text { Sum of first } 10 \text { natural numbers }}{10}\)
    \(=\frac{1+2+3+\ldots \ldots+10}{10}\)
    \(=\frac{55}{10}\)
    \(=5.5\)
  • Question 11
    1 / -0

    The mean of first 10 composite numbers is

    Solution
    The first 10 composite numbers are 4,6,8,9,10,12 , 14,15,16,18
    \(\therefore\) Mean \(=\frac{\text { Sum of first } 10 \text { composite numbers }}{10}\) \(=\frac{4+6+8+9+10+12+14+15+16+18}{10}\)
    \(=\frac{112}{10}\)
    \(=11.2\)
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