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Statistics Test - 20

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Statistics Test - 20
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  • Question 1
    1 / -0
    Find the median for the following data::
    X
    66
    1212
    1818
    2424
    F
    22
    44
    66
    88

    Solution
    X
    Frequency(f)
    Cumulative frequency(cf)
    66
    22
    22
    1212
    44
    2+4=62+4=6
    1818
    66
    6+6=126+6=12
    2424
    88
    12+8=2012+8=20

    n=Σf=20n=\displaystyle\Sigma f=20
    n2=102=10\displaystyle\frac{n}{2}=\frac{10}{2}=10

    n2=102=10\displaystyle\frac{n}{2}=\frac{10}{2}=10

    Therefore,

    the corresponding value of 1010 is 1818.

  • Question 2
    1 / -0
    Find mode for the following data::
    Size
    252-5
    585-8
    8118-11
    111411-14
    Number of shoes
    11
    44
    88
    77

    Solution

    Here frequency of class interval 8118-11 is maximum.

    So, it is the modal class

    Now l l =  lower limit of modal class =8= 8

    f1=f_1 = frequency of modal class =8= 8

    f0=f_0 = frequency of class preceding the modal class =4= 4

    f2=f_2 = frequency of class succeeding the modal class =7= 7

    h=h = class interval width =3= 3

    So, Mode =l+(f1f02f1f0f2)×h=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

    =8+(842×847)×3=8+\left(\displaystyle\frac{8-4}{2\times 8-4-7}\right)\times 3

    =8+(41611)×3=8+\left(\displaystyle\frac{4}{16-11}\right)\times 3

    =8+2.4=8+2.4

    =10.4=10.4

    Mode=10.4=10.4

  • Question 3
    1 / -0
    Find the mode for the following data::
    Students10101414202030306060
    Frequency2222447788
    Solution
    Mode for the discrete series is the variable which has the highest frequency.
    From the given distribution table, the highest frequency is 88 which is associated with the variable 6060
    Therefore, the mode is 6060
  • Question 4
    1 / -0
    Calculate the mode for the following data::
    Daily income
    100200100-200
    200300200-300
    300400300-400
    400500400-500
    500600500-600
    600700600-700
    Workers
    1212
    2323
    4545
    5050
    3030
    1212

    Solution

    Here frequency of class interval 400500400-500 is maximum.

    So, it is the modal class

    Now, l l = lower limit of modal class = 400

    f1=f_1 = frequency of modal class =50= 50

    f0=f_0 = frequency of class preceding the modal class =45= 45

    f2=f_2 = frequency of class succeeding the modal class =30= 30

    h=h = class interval width =100= 100

    So, Mode =l+(f1f02f1f0f2)×h=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

    =400+(50452×504530)×100=400+\left(\displaystyle\frac{50-45}{2\times 50-45-30}\right)\times 100

    =400+(510075)×100=400+\left(\displaystyle\frac{5}{100-75}\right)\times 100

    =400+20=400+20

    =420=420

    Mode=420=420

  • Question 5
    1 / -0
    Find the mode for the following data ::
    Class interval
    020-2
    242-4
    464-6
    686-8
    8108-10
    Frequency
    11
    33
    66
    55
    22

    Solution

    Here frequency of class interval 464-6 is maximum.

    So, it is the modal class

    Now l l = lower limit of modal class =4= 4

    f1=f_1 = frequency of modal class =6= 6

    f0=f_0 = frequency of class preceding the modal class =3= 3

    f2=f_2 = frequency of class succeeding the modal class =5= 5

    h=h = class interval width =2= 2

    So, Mode =l+(f1f02f1f0f2)×h=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

    =4+(632×635)×2=4+\left(\displaystyle\frac{6-3}{2\times 6-3-5}\right)\times 2

    =4+(3128)×2=4+\left(\displaystyle\frac{3}{12-8}\right)\times 2

    =4+1.5=4+1.5

    =5.5=5.5

    Mode=5.5=5.5

  • Question 6
    1 / -0
    The time taken by a group of people to run across the street is given below. Find the median.
    Time(min)
    1010
    2020
    2525
    3030
    4545
    People
    11
    22
    55
    66
    77

    Solution
    Time(min)
    People(f)
    Cumulative frequency(cf)
    1010
    11
    11
    2020
    22
    1+2=31+2=3
    2525
    55
    3+5=83+5=8
    3030
    66
    8+6=148+6=14
    4545
    77
    14+7=2114+7=21

    n=Σf=21n=\displaystyle\Sigma f=21
    n2=212=10.5\displaystyle\frac{n}{2}=\frac{21}{2}=10.5


    n2=212=10.5\displaystyle\frac{n}{2}=\frac{21}{2}=10.5
    Therefore, the corresponding value of time 10.510.5 is 3030min.
  • Question 7
    1 / -0
    Find the approximate value of mode for the following data ::
    Class interval
    7147-14
    142114-21
    212821-28
    283528-35
    354235-42
    Frequency
    22
    44
    33
    1010
    11

    Solution

    Here frequency of class interval 283528-35 is maximum.

    So, it is the modal class

    Now l l = lower limit of modal class =28= 28

    f1=f_1 = frequency of modal class =10= 10

    f0=f_0 = frequency of class preceding the modal class =3= 3

    f2=f_2 = frequency of class succeeding the modal class =1= 1

    h=h = class interval width =7= 7

    So, Mode =l+(f1f02f1f0f2)×h=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

    =28+(1032×1031)×7=28+\left(\displaystyle\frac{10-3}{2\times 10-3-1}\right)\times 7

    =28+(7204)×7=28+\left(\displaystyle\frac{7}{20-4}\right)\times 7

    =28+3.0625=28+3.0625

    =31.06=31.06

    Mode31\sim 31

  • Question 8
    1 / -0
    What is the median class for the following data given below?
    Height(cm)
    0100-10
    102010-20
    203020-30
    304030-40
    405040-50
    506050-60
    Number of players
    33
    66
    22
    55
    1010
    44

    Solution

    To find the median class:

    First add the total number of frequencies.

    Here, f=30f = 30

    So, f2=(Totalnumber offrequency)2\dfrac f2= \dfrac{(Total \quad number \quad of \quad frequency )}{2}

                 =(30)2=\dfrac{ (30 )}{2}

                 =15 = 15

    The cumulative frequency to which f2\dfrac{f}{2} belongs, the value 1515 lies in the class interval 304030-40 from the cummulative frequency table.


    So the median class is 304030-40

  • Question 9
    1 / -0
    Calculate the mode for the following data::
    Score1414161618182020
    Frequency22444488
    Solution
    Mode for the discrete series is the variable which has the highest frequency.
    From the given distribution table, the highest frequency is 88 which is associated with the variable 2020
    Therefore, the mode is 2020
  • Question 10
    1 / -0
    Find the median number of newspaper purchased in a story by 99 customers.
    2,4,5,1,6,8,9,4,72, 4, 5, 1, 6, 8, 9, 4, 7
    Solution
    The median of a set of data is the middlemost number in the set.
    So, first arrange the data in order.
    1,2,4,4,5,6,7,8,91, 2, 4, 4, 5, 6, 7, 8, 9
    The median is 55.
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