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Descriptive Statistics Test 25

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Descriptive Statistics Test 25
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  • Question 1
    1 / -0
    .

    Solution

    $$\Rightarrow$$  One symbol represents $$5$$ balloons.
    $$\Rightarrow$$  Then, $$60$$ balloons represents how many symbols.
    $$\therefore$$  Required symbols $$=\dfrac{60}{5}=12$$
    $$\therefore$$  The number of symbols to be drawn to represent $$60$$ ballons is $$12$$
  • Question 2
    1 / -0
    The pictogram shows the number of pupils who cycle to school. What is the total number of pupils who use cycle for going school?

    Solution
    Number of pupils who use cycle for going school in class 1 $$= 30\times 5 = 150 $$
    Number of pupils who use cycle for going school in class 1 $$= 30\times 7 = 210 $$
    Number of pupils who use cycle for going school in class 1 $$= 30\times 3= 90 $$
    So, Total number of pupils who use cycle for going school $$= 150+210+90 = 450$$

  • Question 3
    1 / -0
    Given above is a bar graph showing the heights of six mountain peaks. Read the above diagram and answer the following:
    Write the ratio of the heights of highest peak and the lowest peak.

    Solution
    From the above bar graph,
    Height of highest peak $$= 8800\ m$$
    Height of lowest peak $$= 6000\  m$$

    Hence the required ratio will be,
    Ratio $$=8800 : 6000$$
              $$ = 22 :15$$

    Hence, option $$A$$ is correct.
  • Question 4
    1 / -0
    Students at a local college were asked how many hours they slept last night. The adjoining chart shows the data . A bar graph of this data will be made on a grid that is 20 units. What scale would be most appropriate for the axis labelled "Number of Students" ?
    Hours of SleepNumber of Students
    614
    726
    828
    915
    More than 106
    Solution
    As there are maximum student are 28 so ,"1unit=2 student" is scale would be most appropriate for the axis labelled "Number of Students".
  • Question 5
    1 / -0
    Standard deviation of the distribution $$1, 3, 5,....., 13$$ will be
    Solution
    Mean of the given numbers is $$\bar x =\dfrac{1+3+5+7+9+11+13}{7}=\dfrac{49}{7}=7$$

    total number of values $$n=7$$

    Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}}$$

    $$\implies SD=\sqrt{\dfrac{(1-7)^2+(3-7)^2+(5-7)^2+(7-7)^2+(9-7)^2+(11-7)^2+(13-7)^2}{7-1}}$$

    $$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{6}}$$

    $$\implies SD=\sqrt{\dfrac{112}{6}}$$

    $$\implies SD=\sqrt{18.6667}=4.32$$

    Therefore the standard deviation for the given values is $$4.32$$
  • Question 6
    1 / -0
    In the following table pass percentage of three schools from the year $$2001$$ to the year $$2006$$ are given. Which school students performance is more consistent?

    $$2001$$$$2002$$$$2003$$$$2004$$$$2005$$$$2006$$
    School (X)$$80$$$$89$$$$79$$$$83$$$$84$$$$65$$
    School (Y)$$92$$$$94$$$$76$$$$75$$$$80$$$$63$$
    School (Z)$$93$$$$97$$$$67$$$$63$$$$70$$$$85$$
    Solution
    Average of $$x=\cfrac { 80+89+79+83+84+65 }{ 6 } =\cfrac { 480 }{ 6 } =80$$
    Average of $$y=\cfrac { 92+94+76+75+80+63 }{ 6 } =\cfrac { 480 }{ 6 } =80$$
    Average of $$z=\cfrac { 93+97+67+63+70+85 }{ 6 } =79.16$$
    Average of x and y are equal and more than z.
    So,their performance is consistent. 
  • Question 7
    1 / -0
    Study the graph carefully and answer the questions given below it. What is the total wheat import during all the given years (in thousand tonnes)?

    Solution
     Total wheat import during all the given years
    $$=3465+1811+2413+4203+7016+5832+2000+2500=29240$$  Thousand tonnes
  • Question 8
    1 / -0
    Find the standard deviation of $$210, 240, 250, 260, 220, 230$$ and $$270$$.
    Solution
    Mean of the given numbers is $$\bar x =\dfrac{210+240+250+260+220+230+270}{7}=\dfrac{1680}{7}=240$$

    total number of values $$n=7$$

    Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$$

    $$\implies SD=\sqrt{\dfrac{(210-240)^2+(240-240)^2+.....+(270-240)^2}{7}}$$

    $$\implies SD=\sqrt{\dfrac{900+0+100+400+400+100+900}{7}}$$

    $$\implies SD=\sqrt{\dfrac{2800}{7}}$$

    $$\implies SD=\sqrt{400}=20$$

    Therefore the standard deviation for the given values is $$20$$
  • Question 9
    1 / -0
    The ages (in years) of a family of 6 members are 1, 5, 12, 15, 38 and 40. The standard deviation is found to be 15.9.  After 10 years the standard deviation is
    Solution
    Then mean of six members=$$\frac{1+5+12+15+38+40}{6}=\frac{111}{6}=18.5$$
    Then Variance $$\alpha ^{2}=\frac{(18.5-1)^{2}+(18.5-5)^{2}+(18.5-12)^{2}+(18.5-15)^{2}+(18.5-38)^{2}+(18.5-40)^{2}}{5}$$
                                   $$=\frac{(17.5)^{2}+(13.5)^{2}+(6.5)^{2}+(3.5)^{2}+(-19.5)^{2}+(21.5)^{2}}{6}$$
                                   $$=\frac{306.25+182.25+42.25+12.25+380.25++462.25}{6}=\frac{1385.50}{6}=230.91$$
    So standard deviation $$\alpha =15.19$$
    Given after 10 year The standard deviation is 15.9
    Then standard deviation remain same
  • Question 10
    1 / -0
    The S.D. of the following freq. dist.
    Class 0-1010-2020-3030-40
    $$f_i$$1342
    Solution
    Class     $$x_i$$   $$f_i$$   $$u_i=\dfrac{x_i-A}{h}$$  $$u_i^2$$        $$f_iu_i$$     $$f_iu_i^2$$
    0-1051-24-24
    10-20153-11-33
    20-302540000
    30-403521122
    Total10-39
    $$S.D.$$$$=h\sqrt { \cfrac { \sum { { f }_{ i }{ u }_{ i }^{ 2 } }  }{ \sum { { f }_{ i } }  } -{ \left( \cfrac { \sum { f } _{ i }{ u }_{ i } }{ \sum { { f }_{ i } }  }  \right)  }^{ 2 } } $$
    $$S.D.=10\sqrt { \cfrac { 9 }{ 10 } -{ \left( \cfrac { -3 }{ 10 }  \right)  }^{ 2 } } =9$$
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