Tag: histograms for non-uniform class widths

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Multiple choice maths bar graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

Strictly monotonical function guarantee inverse mapping as

  1. single valued

  2. multi valued

  3. dual valued

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A strictly monotonic function is always injective (one-to-one). This ensures that for every output in the range, there is exactly one corresponding input in the domain, making the inverse mapping single-valued.

Multiple choice maths bar graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

The statistical data can be represented by diagram using 

  1. Histogram

  2. Frequency polygon

  3. Ogive

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

It is fundamental concept that a statistical data can be represented using plotting graph between different variable, pie chart, histogram, frequency polygon or frequency table or ogives.
Hence, $A,B,C$ are correct choices.

Multiple choice maths bar graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

In a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.
If true then enter $1$ and if false then enter $0$

  1. $1$
  2. $0$

  3. can't determine

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

No, this is not a correct statement.
In a histogram, horizontal axis represents the class intervals whose width is fixed & the varying data is plotted along the y-axis. So in all  the rectangles of a histogram, width remains same & the length changes.
So correct statement is "In a histogram, the area of each rectangle is proportional to the frequency of its class."

Multiple choice maths bar graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

For which of these would you use a histogram to show the data?
$(a)$ The number of letters for different areas in a postman's bag.
$(b)$ The height of competitors in an athletics meet.
$(c)$ The number of cassettes produced by $5$ companies.
$(d)$ The number of passengers boarding trains from $7{:}00$ a.m. to $7{:}00$ p.m. at a station.
Give reasons for each.

  1. $a$
  2. $b$
  3. $c$
  4. $d$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

Histograms are a special form of bar chart where the data represents continuously rather than discrete categories. This means that in a histogram there are no gaps between the columns representing the different categories.

The height of competitors are continuous data. There are no gaps between the data. 
Even the train timings are continuous data.   
Hence, option B and C are correct.

Multiple choice maths interpreting data and graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

Draw the histogram and use it to find the mode for the following frequency distribution.

House - Rent in Rs. per month $4000 - 6000$ $6000 - 8000$ $8000 - 10000$ $10000 - 12000$
Number of families $200$ $240$ $300$ $50$
  1. Rs. $8000$
  2. Rs. $8350$
  3. Rs. $8500$
  4. Rs. $8750$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
max. frequency = $300$
so group is = $8000-10000$
so mode = $L+\dfrac{f _{1}-f _{0}}{2f _{1}-f _{0}-f _{2}}\times w$
$L$= lower class boundary of the modal group
$f _{1}$ = frequency of the modal group
$f _{0}$= frequency of the class preceding or just before the modal class
$f _{2}$ = frequency of the class succeeding or just after the modal class
$w$= group width
mode = $8000+ \dfrac{300-240}{2\times 300-240-50}\times 2000$
mode= $8000+387.096$
mode = $8387.096$
mode = $8350$
Multiple choice maths interpreting data and graphs histograms for non-uniform class widths histograms with unequal class intervals revisiting histograms

Represent the following data by histogram and hence compute mode.

Price of sugar per kg (in Rs.) 18 - 20 20 - 22 22 - 24 24 - 26 26 - 28 Total
Number of weeks 4 8 22 12 6 52

  1. 21.2 Rs.

  2. 22.2 Rs.

  3. 23.2 Rs.

  4. 24.2 Rs.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
max. frequency = $22$
so group is = $22-24$
so mode = $L+\dfrac{f _{1}-f _{0}}{2f _{1}-f _{0}-f _{2}}\times w$
$L$= lower class boundary of the modal group
$f _{1}$ = frequency of the modal group
$f _{0}$= frequency of the class preceding or just before the modal class
$f _{2}$ = frequency of the class succeeding or just after the modal class
$w$= group width
mode = $22+ \dfrac{22-8}{2\times 22-8-12}\times 2$
mode= $22+1.167$
mode = $23.167$
mode = $23.2$