Statistics MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{Line of regression of }}y{\text{ on }}x{\text{ is :}} \cr & y - \overline y = {b_{yx}}\left( {x - \overline x } \right) \cr & \overline y = \frac{{\sum y }}{n}\,;\,\overline x = \frac{{\sum x }}{n} \cr & \Rightarrow \overline y = \frac{{220}}{{10}} = 22\,;\,\overline x = \frac{{130}}{{10}} = 13 \cr & {b_{yx}} = r.\frac{{{\sigma _y}}}{{{\sigma _x}}} \cr & \Rightarrow r = \frac{{n\sum {xy} - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left[ {n\sum {{x^2} - } {{\left( {\sum x } \right)}^2}} \right]\left[ {n\sum {{y^2} - } {{\left( {\sum y } \right)}^2}} \right]} }} \cr & \Rightarrow r = \frac{{10\left( {3467} \right) - \left( {130} \right)\left( {220} \right)}}{{\sqrt {\left[ {\left( {10 \times 2288} \right) - {{130}^2}} \right]\left[ {\left( {10 \times 5506} \right) - {{220}^2}} \right]} }} \cr & \Rightarrow r = 0.962 \cr & {\sigma _y} = \sqrt {\frac{{{{\sum y }^2}}}{n} - {{\left( {\frac{{\sum y }}{n}} \right)}^2}} \cr & \Rightarrow {\sigma _y} = 8.2\,;\,\,{\sigma _x} = 7.73 \cr & \Rightarrow {b_{xy}} = 0.962 \times \frac{{8.2}}{{7.73}} = 1.02 \cr & \Rightarrow {\text{Line of regression of }}y{\text{ on }}x{\text{ is :}} \cr & y - 22 = 1.02\left( {x - 13} \right) \cr & \Rightarrow y = 1.02x + 8.74 \cr} $$

$$\eqalign{ & {\text{Required mean}} \cr & {\text{ = }}\frac{{\left( {a{x_1} + b} \right) + \left( {a{x_2} + b} \right) + ...... + \left( {a{x_n} + b} \right)}}{n} \cr & = \frac{{a\left( {{x_1} + {x_2} + ...... + {x_n}} \right) + nb}}{n} \cr & = a\overline x + b\,\,\,\,\,\,\,\,\left( {\because \frac{{{x_1} + {x_2} + ...... + {x_n}}}{n} = \overline x } \right) \cr} $$

$$\eqalign{ & {\text{Standard deviation}} = \sigma = d\sqrt {\frac{{{n^2} - 1}}{{12}}} \cr & d = {\text{size between each observation}} = 7 \cr & n = {\text{total number of observation}} = 7 \cr & \therefore \,\sigma = 7\sqrt {\frac{{{{\left( 7 \right)}^2} - 1}}{{12}}} \cr & = 7\sqrt {\frac{{49 - 1}}{{12}}} \cr & = 7\sqrt {\frac{{48}}{{12}}} \cr & = 7 \times 2 \cr & = 14 \cr} $$

$$\sigma _x^2 = \frac{{\sum {d_i^2} }}{n}$$    (Here deviations are taken from the mean).
Since $$A$$ and $$B$$ both have $$100$$  consecutive integers, therefore both have same standard deviation and hence the variance.
$$\therefore \,\frac{{{V_A}}}{{{V_B}}} = 1$$   (As $${\sum {d_i^2} }$$  is same in both the cases)

First 50 even natural numbers are 2, 4, 6, . . . . . , 100
Variance $$ = \frac{{\sum {x_i^2} }}{N} - {\left( {\overline x } \right)^2}$$
$$\eqalign{ & \Rightarrow \,\,{\sigma ^2} = \frac{{{2^2} + {4^2} + ..... + {{100}^2}}}{{50}} - {\left( {\frac{{2 + 4 + ..... + 100}}{{50}}} \right)^2} \cr & = \frac{{4\left( {{1^2} + {2^2} + {3^2} + ..... + {{50}^2}} \right)}}{{50}} - {\left( {51} \right)^2} \cr & = 4\left( {\frac{{50 \times 51 \times 101}}{{50 \times 6}}} \right) - {\left( {51} \right)^2} \cr & = 3434 - 2601 \cr & \Rightarrow \,\,{\sigma ^2} = 833 \cr} $$

Mode + 2 Mean = 3 Median
⇒ Mode $$ = 3 \times 22 - 2 \times 21 = 66 - 42 = 24.$$

Mean of $$a, b,$$  8, 5, 10, is 6
$$\eqalign{ & \Rightarrow \,\,\frac{{a + b + 8 + 5 + 10}}{5} = 10 \cr & \Rightarrow \,\,a + b = 7\,\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
Variance of $$a, b,$$  8, 5, 10 is 6.80
$$\eqalign{ & \Rightarrow \,\,\frac{{{{\left( {a - 6} \right)}^2} + {{\left( {b - 6} \right)}^2} + {{\left( {8 - 6} \right)}^2} + {{\left( {5 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}}{5} & = 6.80 \cr & \Rightarrow \,\,{a^2} - 12a + 36 + {\left( {1 - a} \right)^2} + 21 = 34\,\left[ {{\text{using eq}}{\text{. }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow \,\,2{a^2} - 14a + 24 = 0 \cr & \Rightarrow \,\,{a^2} - 7a + 12 = 0 \cr & \Rightarrow \,\,a = 3\,\,{\text{or 4}} \cr & \Rightarrow \,\,b = 4\,\,{\text{or 3}} \cr} $$
∴ The possible values of $$a$$ and $$b$$ are $$a$$ = 3 and $$b$$ = 4 or, $$a$$ = 4 and $$b$$ = 3

Mean deviation is minimum when it is considered about the item, equidistant from the beginning and the end i.e. the median. In this case median is $${\left( {\frac{{101 + 1}}{2}} \right)^{th}}$$   i.e. $${51^{st}}$$  item i.e., $${x_{51}}.$$

If $$s. d. = 0,$$   statements like (A) and (B) can not be given.

$${\text{We construct the following table :}}$$

$${x_i}$$	$$f\left( {{x_i}} \right)$$	$$x_i^2$$	$$f\left( {{x_i}} \right).{x_i}$$	$$f\left( {{x_i}} \right).x_i^2$$
2	$$\frac{1}{3}$$	$$4$$	$$\frac{2}{3}$$	$$\frac{4}{3}$$
3	$$\frac{1}{2}$$	$$9$$	$$\frac{3}{2}$$	$$\frac{9}{2}$$
11	$$\frac{1}{6}$$	$$121$$	$$\frac{11}{6}$$	$$\frac{121}{6}$$
	$$1$$		$$4$$	$$26$$

$$\eqalign{ & \therefore \,{\text{Variance}}\,{\sigma ^2} = \frac{{\sum {f\left( {{x_i}} \right).x_i^2} }}{{\sum {f\left( {{x_i}} \right)} }} - {\left( {\frac{{\sum {f\left( {{x_i}} \right).{x_i}} }}{{\sum {f\left( {{x_i}} \right)} }}} \right)^2} \cr & = \frac{{26}}{1} - {\left( {\frac{4}{1}} \right)^2} \cr & = 26 - 16 \cr & = 10 \cr} $$