Tag: revisiting histograms

Questions Related to revisiting histograms

Local histogram processing deals with

  1. whole image

  2. slices of image

  3. centre of image

  4. none of these


Correct Option: B
Explanation:

In local histogram we deal with slice of images

Strictly monotonical function guarantee inverse mapping as

  1. single valued

  2. multi valued

  3. dual valued

  4. none of these


Correct Option: A

Histogram is technique processed in

  1. intensity domain

  2. frequency domain

  3. spatial domain

  4. none of these


Correct Option: C
Explanation:

Well known from application of histogram

The statistical data can be represented by diagram using 

  1. Histogram

  2. Frequency polygon

  3. Ogive

  4. None of these


Correct Option: A,B,C
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.

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


Correct Option: B
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."

The width of a rectangle in a histogram represents ___________ of the class.

  1. frequency

  2. range

  3. class limit

  4. upper limit


Correct Option: A
Explanation:

The height of the rectangle in the histogram represents the frequency.

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. $d$


Correct Option: B,D
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.

An ogive curve is 

  1. Histogram

  2. Frequency polygon

  3. Cumulative frequency

  4. All the above


Correct Option: A

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. $8750$


Correct Option: B
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 = $8350$

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.


Correct Option: C
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 = $23.2$