Correct2
Incorrect-0.5
If the sum of 11 consecutive natural numbers is 2761, then the middle number is:
249
250
251
252
Let the first natural number be x
According to the question
x+x+1x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10=2761
=> 11x + 55 = 2761
x = {tex} \frac{2761 - 55}{11} = 246 {/tex}
Middle number = x + 5 = 246 + 5 = 251
Correct2
Incorrect-0.5
The mean weight of 9 items is 15. If one more item is added to the series, the mean 16. The value of 10th item is
35
30
25
20
Total weight of 9 items = 15 x 9 = 135
and total weigth of 10 items = 16 X 10 = 160
Therefore, weight of 10 items = 160 - 135 = 25
Correct2
Incorrect-0.5
The mean of 12 items is {tex} \bar{X} {/tex} . If the first term is increased by 1, second by 2 and so on, then the new mean is
{tex} \overline{X}+n {/tex}
{tex} \overline{X}+\frac{n}{2} {/tex}
{tex} \overline{X}+\frac{n+1}{2} {/tex}
none of these
Correct2
Incorrect-0.5
If the mean of n observation {tex} 1^{2},\ 2^{2},\ 3^{2},........n^{2} {/tex}, is {tex} \frac{46}{11}n {/tex}, then n is equal to
11
12
23
22
Correct2
Incorrect-0.5
The value of mean, median and mode coincides, then the distribution is
Positive skewness
Symmetrical distribution
Negative skewness
All of these
If mean, median and mode coincides, then there is a symmetrical distribution.
Correct2
Incorrect-0.5
In a class of 100 students, the average amount of pocket money is Rs.35 per student. If the average is Rs.25 for girls and Rs.50 for boys, then the number of girls in the class is
20
40
60
80
Let the number of girls in the class = y
Therefore, Number of boys in the class = 100 - y
Now, {tex} \overline{x}_{1}=25,n_{1}=y,\overline{x}_{2}=50,n_{2}=100-y\ {/tex}
and {tex} \overline{x}=35,n_{1}+n_{2}=100 {/tex}
{tex} \therefore\ 35 = \frac{25 \times y+50 \times \left(100-y\right) }{100} {/tex}
=> 3500 = 25y + 5000 - 50y
=> 25y = 1500 => y = 60
=> Number of girls in the class = 60
Correct2
Incorrect-0.5
For a series the value of mean deviation is 15, the most likely value of its quartile deviation is
12.5
11.6
13
9.7
Since,
{tex} MD = \frac{4}{5}\sigma,\ QD=\frac{2}{3}\sigma {/tex}
{tex} \therefore\ \frac{MD}{QD} = \frac{6}{5} {/tex}
=> {tex} QD = \frac{5}{6} \left(MD\right)=\frac{5}{6} \left(15\right)=12.5 {/tex}
Correct2
Incorrect-0.5
If {tex} \overline{x} {/tex} is the arithmetic mean of n independent variates {tex} x_{1},\ x_{2},\ x_{3}.....,\ x_{n} {/tex} each of the standard deviation {tex} \sigma {/tex}, the variance {tex} \overline{x} {/tex} is
{tex} \frac{\sigma^{2}}{n} {/tex}
{tex} \frac{n\sigma^{2}}{2} {/tex}
{tex} \frac{\left(n+1\right)\sigma^{2}}{3} {/tex}
None of these
We have, {tex} \overline{x} = \frac{1}{n}\sum^{n}_{i=1}x_{i} {/tex}
Therefore, {tex} var \left(\overline{x}\right) = \frac{1}{n^{2}}\left[ \sum^{n}_{i=1} var \left(x_{j}\right)+2\sum^{n}_{i \neq j} cov \left(x_{i},\ x_{j}\right) \right] {/tex}
Correct2
Incorrect-0.5
Coefficient of skewness for the values median {tex} = 18.8,\ Q_{1} = 14.6,\ Q_{3} = 25.2 {/tex} is
0.2
0.5
0.7
None of these
Coefficient of skewness = {tex} \frac{Q_{3}-Q_{1}-2 \left(median\right) }{Q_{3}-Q_{1}} {/tex}
= {tex} \frac{25.2+14.6-2 \left(18.8\right) }{25.2-14.6} = \frac{2.2}{10.6} = 0.20 {/tex}
Correct2
Incorrect-0.5
The 7th percentile is equal to
7th decile
{tex} Q_{3} {/tex}
6th decile
None of these
7th decile {tex} D_{7} = \frac{7n}{10} {/tex} .. (i)
and 7th percentile, {tex} P_{10} = \frac{7n}{100} {/tex} .. (ii)
From equations (i) and (ii), we get
{tex} D_{7}\neq P_{70} {/tex}
Correct2
Incorrect-0.5
Consider the following statements:
1. The values of median and mods can be determined graphically.
2. Mean, median and mode have the same unit.
3. Range is the best measure of dispersion
which of these is/are correct?
(1) alone
(2) alone
Both (2) and (3)
None of these
It is true that median and mode can be determined graphically.
Correct2
Incorrect-0.5
In a factory daily wages of 7 employees are Rs.50, Rs.60, Rs.70, Rs.80, Rs.90, Rs.100 and Rs.110. Calculate the arithmetic mean (AM)
80
70
90
60
Arithmetic mean (AM) = {tex} \frac{50+60+70+80+90+100+110}{7} {/tex}
={tex} \frac{560}{7} {/tex} = 80
Correct2
Incorrect-0.5
Calculate the mean of the following marks secured by 40 students
| Marks: | 35 | 40 | 45 | 50 | 65 | 70 |
| Number of students: | 4 | 3 | 2 | 16 | 8 | 7 |
74
54
84
64
| Marks(x) | Frequency(f) | fx |
| 35 | 4 | 140 |
| 40 | 3 | 120 |
| 45 | 2 | 90 |
| 50 | 16 | 800 |
| 65 | 8 | 520 |
| 40 | 7 | 490 |
| {tex} \sum x=40 {/tex} | {tex} \sum fx =2160{/tex} |
Correct2
Incorrect-0.5
Calculate the mean for the following frequency distribution.
| x: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| f: | 6 | 8 | 9 | 10 | 7 | 5 | 15 |
54.32
62.24
48.16
42.14
| x | f | m | {tex} x' = \frac{m-A}{i} {/tex} | fx' |
| 10-20 | 6 | 15 | -3 | -18 |
| 20-30 | 8 | 25 | -2 | -16 |
| 30-40 | 9 | 35 | -1 | -9 |
| 40-50 | 10 | 45 | 0 | 0 |
| 50-60 | 7 | 55 | 1 | 7 |
| 60-70 | 5 | 65 | 2 | 10 |
| 70-80 | 15 | 75 | 3 | 45 |
| {tex} \sum =60 {/tex} | {tex} \sum fx'=19 {/tex} |
Correct2
Incorrect-0.5
Find the median for the following data 15, 24, 18, 9, 23, 6, 19
24
6
19
18
Arrange the given data in ascending order. The data can be written as
6. 9, 15, 18, 19, 23, 24
The middle No. is 18.
Hence 18 is the median.
Correct2
Incorrect-0.5
Find the median for the following data 42, 24, 51, 33, 15, 60
37.5
38.5
40
40.5
Arrange the given data in ascending order. The data can be written as
15, 24, 33, 42, 51, 60
Here the number of data is even number. So take the two middle data and find the average.
Median = {tex} \frac{33+42}{2}=\frac{75}{2} = 37.5{/tex}
Correct2
Incorrect-0.5
Find the standard deviation for the given data.
The weight odf 5 boys (in kg.) are given below:
45, 40, 36, 34, 65
13.25
11.15
14.35
12.25
| x | {tex} d=x-\overline{x} {/tex} | {tex} d^{2} {/tex} |
| 45 | 45-44=1 | 1 |
| 40 | 40-44=-4 | 16 |
| 36 | 36-44=-8 | 64 |
| 34 | 34-44=-10 | 100 |
| 65 | 65-44=21 | 441 |
| {tex} \sum d^{2} {/tex} | 622 |
Correct2
Incorrect-0.5
When dice is thrown, find the probability to get the sum of the numbers is 5.
{tex} \frac{1}{4} {/tex}
{tex} \frac{1}{3} {/tex}
{tex} \frac{1}{2} {/tex}
{tex} \frac{1}{5} {/tex}
Number of occurrence =
{tex} \{ \left(1, 1\right) \left(1, 2\right) \left(1, 3\right) \left(1, 4\right) \left(1, 5\right) \left(1, 6\right) {/tex} {tex} \left(2, 1\right) \left(2, 2\right) \left(2, 3\right) \left(2,4\right) \left(2, 5\right) \left(2, 6\right) {/tex} {tex} \left(3, 1\right) \left(3, 2\right) \left(3, 3\right) \left(3, 4\right) \left(3, 5\right) \left(3, 6\right) {/tex} {tex} \left(4, 1\right) \left(4, 2\right) \left(4, 3\right) \left(4, 4\right) \left(4, 5\right) \left(4, 6\right) {/tex} {tex} \left(5, 1\right) \left(5, 2\right) \left(5, 3\right) \left(5, 4\right) \left(5, 5\right) \left(5, 6\right) {/tex} {tex} \left(6, 1\right) \left(6, 2\right) \left(6, 3\right) \left(6, 4\right) \left(6, 5\right) \left(6, 6\right) \} {/tex}
n(S) = 36
Number of events to get total of 5 is {tex} \{ \left(1, 4\right) \left(2, 3\right) \left(3, 2\right) \left(4, 1\right) \} {/tex}
n(E) =4
p(E) = {tex} \frac{n \left(E\right) }{n \left(S\right) }=\frac{4}{36}=\frac{1}{9} {/tex}
Note: Probability to get either head or tail when the coin is tossed is {tex} \frac{1}{2} {/tex}
Correct2
Incorrect-0.5
When three coins are tossed. Find the probability to get at most one head.
{tex} \frac{2}{8} {/tex}
{tex} \frac{5}{8} {/tex}
{tex} \frac{7}{8} {/tex}
{tex} \frac{3}{8} {/tex}
When the three coins are tossed.
S = {tex} \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \} {/tex}
n(S) = 8
n(E) = {tex} \{ HTT, THT,TTH \} = 3 {/tex}
p(E) = {tex} \frac{n \left(E\right) }{n \left(S\right) }=\frac{3}{8} {/tex}