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CGL Tier 2 Paper 3 JSO |Questions with Detailed Explanation|

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CGL Tier 2 Paper 3 JSO |Questions with Detailed Explanation|

CGL Tier 2 Paper 3 JSO |Questions with Detailed Explanation|

Q1. What is the purpose of a summary table?

(a)This is the only way to present categorical data in numerical form.

(b)To see differences between or among categories.

(c)To sum the values of responses to a survey.

(d)To list data to create a bar or pie chart.

Answer: (b)

While the summary table may be used for this purpose, the primary purpose is to allow the analyst to see differences between or among categories.

Q2. A graphical representation of a frequency distribution is called a_______.

(a)scatter diagram

(b)stem-and-leaf plot

(c)time-series plot

(d)histogram

Answer: (d)

A stem-and-leaf plot does group the data but does not explicitly give the frequencies.

Q3. You have a summary table and a simple bar chart-like the ones at the beginning of the chapter indicating where customers prefer to do their banking. How could you enhance the bar chart to provide both visual and actual information?

(a)Use vertical lines on the bar chart to show the values more precisely.

(b)Add values to the bar chart like what is commonly done on a pie chart.

(c)Only the summary table can show the actual values for the data.

(d)The bar chart and summary table must be presented together in order to represent this data.

Answer: (b)

The values from the summary table can be added to the bar chart.

Q4. It might be said that the stem-and-leaf display is really a quick and easy way of creating a rudimentary chart or diagram for numerical data. If so, which chart is used to describe categorical data does it most closely resemble?

(a)The stem-and-leaf display most closely resembles a rudimentary bar chart.

(b)The stem-and-leaf display most closely resembles a rudimentary pie chart.

(c)The stem-and-leaf display most closely resembles a rudimentary Pareto chart.

(d)The stem-and-leaf display does not resemble any of the above charts or diagrams.

Answer: (a)

While the stem-and-leaf display does not exactly resemble any of these charts, it most closely resembles a bar chart.

Q5. The width of a class interval in a frequency distribution -or bar chart will be approximately equal to the range of the data divided by the ______________ .

(a)average of the data set

(b)number of class intervals

(c)highest value in the data set

(d)lowest value in the data set

Answer: (b)

The width of a class interval to be the range divided by the number of classes.

Q6. White dividing each entry in a data by a non zero number, a the arithmetic mean of the new data

(a)Is multiplied by a

(b)Does not change

(c)Is divided by a

(d)Is diminished by a

Answer: (c)

Since the values of the variable X are multiplied (or dividing) by a constant , the arithmetic mean of the new observations can be obtained by multiplying (or dividing) the initial arithmetic mean by the same constant.

Therefore, when dividing each entry in a data by non zero number, a, the arithmetic mean of new data is divided by a.

Q7. The sum of deviations taken from the actual arithmetic mean is

(a)Zero

(b)Two

(c)Negative

(d)Infinite

Answer: (a)

Since Algebraic sum of the deviations of a set of values from their arithmetic mean is zero. If  is the frequency distribution, then  being the mean of the distribution.Therefore the sum of deviations taken from the actual arithmetic mean is zero.

Q8. The mean age of 40 students is 16 years and the mean age of another group of 60 students is 20 years. The mean age of all the 100 students is (in years)

(a)16.8

(b)18

(c)18.4

(d)18.8

[showhide type="links8" more_text="Show Answer" less_text="Hide Answer"]

Answer: (c)

Here 

Combined mean of all the 100 students will be

Q9. The point of intersection of two ogives, if a frequency distribution with median is 30 and the total frequency is 100, is

(a)35, 50

(b)30, 50

(c)40, 50

(d)50, 30

Answer: (b)

In locating median for the grouped data, less than and more than ‘ogive’ have been considered. The intersecting point of the two curves, when projected on X- axis, gives median value, therefore point of intersection of two ogives is 30, 50.

Q10. Which is the correct relation for a moderately asymmetrical distribution?

(a)A.M. < G.M. < H.M.

(b)A.M. > G.M. > H.M.

(c)G.M > A.M. > H.M.

(d)G.M < A.M. < H.M

Answer: (b)

For asymmetrical distribution, the relation between A.M. , G.M. , and H.M. , is A.M. > G. M. > H.M.