(A) lower limits of the classes

(B) upper limits of the classes

(C) midpoints of the classes

(D) frequencies of the class marks.

Answer:  (C)

Explanation: We know that d_i = x_i – a_i. i.ed_i’s are the deviations from the midpoints of the classes.

Q2. While computing mean of the grouped data, we assume that the frequencies are:

(A) evenly distributed over all the classes                                         

(B) centered at the class marks of the classes

(C) centered at the upper limits of the classes

(D) centered at the lower limits of the classes

Answer:  (B)

Explanation: In computing the mean of grouped data, the frequencies are centred at the class marks of the classes

(A) 0

(B) – 1

(C) 1

(D) 2

Answer:  (A)

Explanation:

Q4. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency of a grouped data gives its:

(A) Mean

(B) Median

(C) Mode

(D) All of these

Answer:  (B)

Explanation: Since the intersection point of less than type ogive and more than ogive gives the median on the abscissa.

Q5. 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 median class and modal class is:

(A) 15

(B) 25

(C) 30

(D) 35

Answer:  (B)

Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Now N/2 = 66/2 = 33 which lies in the interval 10 – 15.Therefore lower limit of the median class is 10.

The highest frequency is 20 which lies in the interval 15 – 20. Therefore, lower limit of modal class is 15.

Hence required sum is 10 + 15 = 25

Q6.  If the arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x = ?

(A) 1

(B) 2

(C) 6

(D) 4

Answer:  (D)

Explanation:

According to question

Q7. If the mean of first n natural numbers is 5n/9, then n =?

(A) 6

(B) 7

(C) 9

(D) 10

Answer:  (C)

Explanation:

But according to question,

Q8. If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by:

(A) 2

(B) 1.5

(C) 1

(D) 0.5

Answer:  (D)

Explanation: We have

30, 34, 35, 36, 37, 38, 39, 40

The data has 8 observations, so there are two middle terms, 4^th and 5^th term i.e. 36 and 37.

The median is the mean of both these terms.

Median = (36 + 37)/2

Median = 36.5

When 35 is removed from given data as 30, 35, 36, 37, 38, 39, 40 then the number of observations becomes 7.

Now the median is the middle most i.e 4^th term which is equal to 37.

Therefore median is increased by 37 – 36.5 = 0.5

Q9. The Median when it is given that mode and mean are 8 and 9 respectively, is:

 (A) 8.57

(B) 8.67

 (C) 8.97

(D) 9.24

Answer:  (B)

Explanation: By Empirical formula:

Mode = 3median – 2 mean

8 = 3medain – 2 X 9

8 = 3median – 18

3median = 8 + 18

Median = 26/3

Median = 8.67

(A) 3

(B) 4

(C) 5

(D) 6

Answer:  (D)

Explanation: According to question,

Q11. In a hospital, weights of new born babies were recorded, for one month. Data is as shown:

Weight of new born baby (in kg)	1.4 – 1.8	1.8 – 2.2	2.2 – 2.6	2.6 – 3.0
No of babies	3	15	6	1

Then the median weight is:

(A) 2kg

(B) 2.03kg

(C) 2.05 kg

(D) 2.08 kg

Answer:  (C)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	Cumulative Frequency (cf)
1.4-1.8	3	1.6	3
1.8-2.2	15	2	18
2.2-2.6	6	2.4	24
2.6-3.0	1	2.8	25

Since N/2 = 25/2 = 12.5

12.5 is near to cumulative frequency value 18

So median class interval is 1.8 – 2.2

∴Median = l + [(N/2 – cf)/f]/h

Here

Hence median weight is 2.05 kg.

Q12. In a small scale industry, salaries of employees are given in the following distribution table:

Salary (in Rs.)	4000 – 5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000
Number of employees	20	60	100	50	80	90

Then the mean salary of the employee is:

(A) Rs. 7350

(B) Rs.  7400

(C) Rs. 7450

(D) Rs. 7500

Answer: (C)

Explanation:

Therefore mean is:

Q13.  For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are:

Number of days	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40	TOTAL
Total Number of students	15	16	x	8	y	8	6	4	70

(A) x = 4 and y = 3

(B) x = 7 and y = 7

(C) x = 3 and y = 4

(D) x = 7 and y = 6

Answer:  (D)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi
0-5	15	2.5	37.5
5 – 10	16	7.5	120
10 – 15	x	12.5	12.5x
15 – 20	8	17.5	140
20 – 25	y	22.5	22.5y
25 -30	8	27.5	220
30 – 35	6	32.5	195
35 – 40	4	37.5	150
TOTAL	70		12.5x+22.5y+862.5

mean = (12.5x + 22.5y + 862.5)/70

⇒ 15.5 = (12.5x +22.5y + 862.5)/70

⇒ 15.5 X 70 = 12.5x +22.5y + 862.5

⇒ 12.5x + 22.5y = 222.5

⇒ 125x + 225y = 2225

⇒ 5x + 9y = 89                        …..(i)

Also,

x + y + 57 = 70

x + y = 13         ……(ii)

Multiplying equation (ii) by 5 and then subtracting from (i) as,

Substituting the value of y in equation (ii), we get

x + y = 13

⇒ x + 6 = 13

⇒ x = 7

Hence x = 7 and y = 6

Q14. Pocket expenses of a class in a college are shown in the following frequency distribution:

Pocket expenses	0-200	200-400	400-600	600-800	800-1000	1000-1200	1200-1400
Number of students	33	74	170	88	76	44	25

Then the median for the above data is:

(A) 485.07

(B) 486.01

(C) 487.06

(D) 489.03

Answer:  (C)

Explanation:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi	cf
0-200	33	100	3300	33
200-400	74	300	22200	107
400-600	170	500	85000	277
600-800	88	700	61600	365
800-1000	76	900	68400	441
1000-1200	44	1100	48400	485
1200-1400	25	1300	32500	510
	510		321400

Since N/2 = 510/2 = 255

255 is near to cumulative frequency value 277.

So median class interval is 400-600

Here,

l = 400

N/2 = 255

cf = 107

f = 170

h = 100

Therefore,

Answer:  (B)

Explanation: We have

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