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Statistics Test - 35

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Statistics Test - 35
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  • Question 1
    1 / -0
    How many students are over $$170\:cm$$ tall?

    Solution
    From the histogram:
    Height (in cm)No. of students 
    $$150-155 $$ $$5 $$
    $$155-160$$ $$1$$
    $$160-165$$ $$ 5$$
    $$ 165-170$$ $$5$$
    $$ 170-175$$ $$ 9$$
    $$ 175-180$$ $$5$$
    Then, number of students whose height is over $$170 cm$$
    $$=n(170)+n(175)$$
    $$=9+5$$
    $$=14$$.

    Hence, $$14$$ students are over $$170 cm$$ tall.
    Therefore, option $$A$$ is correct.
  • Question 2
    1 / -0

    Directions For Questions

    Mr. Daniel surveyed his employees on the time it takes them to get to work. The result are shown in the histogram.

    ...view full instructions

    How many employees get to work in more than $$100$$ minutes?

    Solution
    From the histogram, we can observe the class width of:
    $$100-120$$ minutes $$= 4$$ employees.
    Therefore, number of employees get into work in more than $$100$$ minutes $$= 4$$.
    So, $$4$$ employees get into work more than $$100$$ minutes.

    Hence, option $$C$$ is correct.
  • Question 3
    1 / -0
    Convert the following data into grouped frequency table. Find the total frequency:
    $$2, 3, 4, 12, 11, 2, 4, 3, 15, 15, 16, 2, 3, 20, 20$$
    Solution
    Arrange the data in ascending order: $$2, 2, 2, 3, 3, 3, 4, 4, 11, 12, 15, 15, 16, 20, 20$$
    Range $$=$$ largest number $$-$$ smallest number
    Range $$= 20 - 2 = 18$$
    Let us form $$4$$ groups from the data, class width $$= \dfrac{\text{range}}{4} = \dfrac{18}{4} = 4.5$$ round off the number, we get $$5$$.
    Form a table,
    Class interval   
    Frequency
    $$0-5$$
    $$8$$
    $$6-11$$
    $$1$$
    $$12-17$$
    $$4$$
    $$18-23$$
    $$2$$
    So, the total frequency $$=8+1+4+2=15$$.
  • Question 4
    1 / -0

    Directions For Questions

    The number of ducks at each pond in the Mumbai city is shown in the histogram.

    ...view full instructions

    In which class interval number of ducks is equal to $$20$$ ponds?

    Solution
    From the histogram, the class width of number of ducks:
     $$0 - 19 = 30$$ number of ponds.
     $$20 - 39 = 20$$ number of ponds.
     $$40 - 59 = 25$$ number of ponds.
     $$60 - 79 = 10$$ number of ponds.
     $$80 - 99 = 20$$ number of ponds.
     $$100 - 119 = 15$$ number of ponds.

    Therefore, $$20-39$$ and $$80-99$$ is the two class interval equal to $$20$$ ponds.

    Hence, option $$C$$ is correct.
  • Question 5
    1 / -0
    $$70$$ number of student's height are measured in cm as shown in the histogram. How many students have heights more than $$180$$ cm?

    Solution
    The given histogram represents the height of $$70$$ students in $$cm$$.

    For finding the number of students having a height of more than $$180\ cm$$, we need to add the heights of the bars representing the bars $$180-200,\ 200-220,\ 220-240$$
    The height of the bar, representing the number of students is:
    $$180-200\rightarrow 10$$
    $$200-220\rightarrow 15$$
    $$220-240\rightarrow 7$$

    Therefore, number of students having height more than $$180\ cm= 10 + 15 + 7$$
    $$ = 32$$

    Hence, $$32$$ students have height more than $$180\ cm$$.
  • Question 6
    1 / -0

    Directions For Questions

    Mr. Daniel surveyed his employees on the time it takes them to get to work. The result are shown in the histogram.

    ...view full instructions

    How many employees get to work in less than $$20$$ minutes?

    Solution
    From the histogram, we can observe the class width of:
    $$0-20$$ minutes $$= 4$$ employees.

    Therefore, the number of employees get into work in less than $$20$$ minutes $$= 4$$.
    So, $$4$$ employees get into work less than $$20$$ minutes.

    Hence, option $$A$$ is correct.
  • Question 7
    1 / -0

    Directions For Questions

    Mr. Daniel surveyed his employees on the time it takes them to get to work. The result are shown in the histogram.

    ...view full instructions

    How many employees get to work in less than $$60$$ minutes?

    Solution
    From the histogram, we can observe the class width of:
    $$0-20$$ minutes $$= 4$$ employees,
    $$20-40$$ minutes $$= 6$$ employees,
    $$40-60$$ minutes $$ = 10$$ employees.

    Therefore, number of employees get into work in less than $$60$$ minutes $$= 4 + 6 + 10 = 20$$.

    So, $$20$$ employees get into work less than $$60$$ minutes.

    Hence, option $$D$$ is correct.
  • Question 8
    1 / -0

    Directions For Questions

    The number of ducks at each pond in the Mumbai city is shown in the histogram.

    ...view full instructions

    How many number of ducks are there in less than $$60$$ ponds?

    Solution
    From the histogram, the class width gives the number of ducks:
     $$0 - 19 = 30$$ number of ponds.
     $$20 - 39 = 20$$ number of ponds.
     $$40 - 59 = 25$$ number of ponds.
     $$60 - 79 = 10$$ number of ponds.
     $$80 - 99 = 20$$ number of ponds.
     $$100 - 119 = 15$$ number of ponds.

    Therefore, $$30+20+25=59$$ number of ducks are there in less than $$60$$ ponds.

    Hence, option $$A$$ is correct.
  • Question 9
    1 / -0

    Directions For Questions

    $$70$$ number of student's heights are measured in cm as shown in the histogram.

    ...view full instructions

    How many students have heights less than $$180$$ cm?

    Solution
    From the histogram, the frequency associated with the class width is as follows:
    $$120 - 140$$ is $$4$$
    $$140 - 160$$ is $$10$$
    $$160 - 180$$ is $$24$$
    $$180 - 200$$ is $$10$$
    $$200 - 220$$ is $$15$$
    $$220 - 240$$ is $$7$$.

    Therefore, number of students who have heights less than $$180$$ cm $$= 24 + 10 + 4 = 38$$.

    Hence, option $$C$$ is correct.
  • Question 10
    1 / -0
    Convert the following data into grouped frequency table with class interval length of $$5$$. Find the third interval frequency:
    $$12, 12, 11, 12, 2, 3, 3, 3, 4, 4, 2, 2, 5, 6, 7, 4, 7, 8, 15, 15, 20, 20$$
    Solution
    Arrange the data in increasing order: $$2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 7, 7, 8, 11, 12, 12, 12, 15, 15, 20, 20$$

    Form the required frequency distribution table with interval of $$5$$

     Class interval Frequency
     $$0-5$$ $$10$$
     $$6-11$$ $$5$$
     $$12-17$$ $$5$$
     $$18-23$$ $$2$$
    Arrange the data in order: $$2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 7, 7, 8, 11, 12, 12, 12, 15, 15, 20, 20$$

    So, the third interval frequency is $$5$$.

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