In the formula \(\bar{x}=a+\dfrac{\sum f_i d_i}{\sum f_i}\), for finding the mean of grouped data, \(d_i\)'s are deviations from \(a\) of
lower limits of the classes
upper limits of the classes
mid-points (class marks) of the classes
frequencies of the class marks
While computing mean of grouped data, we assume that the frequencies are
evenly distributed over all the classes
centred at the class marks of the classes
centred at the upper limits of the classes
centred at the lower limits of the classes
If \(x_i\) are the class marks and \(f_i\) the corresponding frequencies with mean \(\bar{x}\), then \(\sum f_i(x_i-\bar{x})\) equals
0
-1
1
2
In the formula \(\bar{x}=a+h\,\dfrac{\sum f_i u_i}{\sum f_i}\) for mean of grouped data, \(u_i=\)
\(\dfrac{x_i+a}{h}\)
\(h(x_i-a)\)
\(\dfrac{x_i-a}{h}\)
\(\dfrac{a-x_i}{h}\)
The abscissa of the intersection point of the less-than and more-than cumulative frequency curves of a grouped data gives its
mean
median
mode
all the three above
For the following distribution:
| Class | 0–5 | 5–10 | 10–15 | 15–20 | 20–25 |
|---|---|---|---|---|---|
| Frequency | 10 | 15 | 12 | 20 | 9 |
the sum of lower limits of the median class and modal class is
15
25
30
35
Consider the distribution:
| Class | 0–5 | 6–11 | 12–17 | 18–23 | 24–29 |
|---|---|---|---|---|---|
| Frequency | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is
17
17.5
18
18.5
For the following distribution:
| Marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 | Below 60 |
|---|---|---|---|---|---|---|
| No. of students | 3 | 12 | 27 | 57 | 75 | 80 |
The modal class is
10–20
20–30
30–40
50–60
Consider the data:
| Class | 65–85 | 85–105 | 105–125 | 125–145 | 145–165 | 165–185 | 185–205 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is
0
19
20
38
Times (s) taken by 150 athletes in a 110 m hurdle race:
| Class | 13.8–14 | 14–14.2 | 14.2–14.4 | 14.4–14.6 | 14.6–14.8 | 14.8–15 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 4 | 5 | 71 | 48 | 20 |
The number who finished in less than 14.6 s is
11
71
82
130
Consider:
| Marks obtained | ≥0 | ≥10 | ≥20 | ≥30 | ≥40 | ≥50 |
|---|---|---|---|---|---|---|
| No. of students | 63 | 58 | 55 | 51 | 48 | 42 |
The frequency of the class 30–40 is
3
4
48
51
If an event cannot occur, its probability is
1
\(\dfrac{3}{4}\)
\(\dfrac{1}{2}\)
0
Which of the following cannot be the probability of an event?
\(\dfrac{1}{3}\)
0.1
3%
\(\dfrac{17}{16}\)
An event is very unlikely to happen. Its probability is closest to
0.0001
0.001
0.01
0.1
If the probability of an event is \(p\), the probability of its complementary event is
\(p-1\)
\(p\)
\(1-p\)
\(1-\dfrac{1}{p}\)
The probability expressed as a percentage of a particular occurrence can never be
less than 100
less than 0
greater than 1
anything but a whole number
If \(P(A)\) denotes the probability of an event A, then
\(P(A)<0\)
\(P(A)>1\)
\(0\le P(A)\le 1\)
\(-1\le P(A)\le 1\)
A card is selected from a deck of 52 cards. The probability that it is a red face card is
\(\dfrac{3}{26}\)
\(\dfrac{3}{13}\)
\(\dfrac{2}{13}\)
\(\dfrac{1}{2}\)
The probability that a non-leap year selected at random will contain 53 Sundays is
\(\dfrac{1}{7}\)
\(\dfrac{2}{7}\)
\(\dfrac{3}{7}\)
\(\dfrac{5}{7}\)
When a die is thrown, the probability of getting an odd number less than 3 is
\(\dfrac{1}{6}\)
\(\dfrac{1}{3}\)
\(\dfrac{1}{2}\)
0
A card is drawn from a deck of 52 cards. Event \(E\): “card is not an ace of hearts”. The number of outcomes favourable to \(E\) is
4
13
48
51
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
7
14
21
28
A girl finds the probability of winning the first prize in a lottery to be 0.08. If 6000 tickets are sold, how many tickets has she bought?
40
240
480
750
One ticket is drawn at random from a bag of tickets numbered 1 to 40. The probability that the selected ticket is a multiple of 5 is
\(\dfrac{1}{5}\)
\(\dfrac{3}{5}\)
\(\dfrac{4}{5}\)
\(\dfrac{1}{3}\)
Someone is asked to take a number from 1 to 100. The probability that it is a prime is
\(\dfrac{1}{5}\)
\(\dfrac{6}{25}\)
\(\dfrac{1}{4}\)
\(\dfrac{13}{50}\)
A class has 23 students: 4 from house A, 8 from B, 5 from C, 2 from D and the rest from E. One student is selected at random to be the monitor. The probability that the selected student is not from A, B and C is
\(\dfrac{4}{23}\)
\(\dfrac{6}{23}\)
\(\dfrac{8}{23}\)
\(\dfrac{17}{23}\)