Solution:

Now,
Sum of a numbers = 12 + 15 + 18 + 19 + 15 + 17 + 16 + 10 + 13 + 15
= 150
number = 10
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or,Mean (\(\overline{X}\)) = \(\frac{150}{10}\)
\(\therefore\) Mean (\(\overline{X}\)) = 15

Solution:

Sum of given terms (\(\sum\)x) = 66 mph + 57 mph + 71 mph + 54 mph + 69 mph + 58 mph = 375

numbers of terms (n) = 6

We have,

Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)

= \(\frac{375}{6}\)

= 62.5

The mean driving speed is 62.5 mph.

Solution:

Sum of given terms (\(\sum\)x) = 89+ 73+ 84+ 91+ 87+ 77+ 94 = 515

numbers of terms (n) =7

We have,

Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)

= \(\frac{515}{7}\)

= 73.57

Hence, The mean test score is 73.57

Solution:

Sum of given terms (\(\sum\)x) = $1.79+ $1.61+ $1.96+ $2.08 = $7.44

numbers of terms (n) = 4

We have,

Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)

= \(\frac{7.44}{7}\)

= 1.86

Hence, The mean gasoline price is $1.86

Solution:

Now,
Sum of numbers = 3 + 7 + 10 + 15 + x
= 35 + x
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, \(\frac{12}{1}\)= \(\frac{35 + x}{n}\)
or, 60 = 35 + x
or, x = 60 - 35
or, x = 25
\(\therefore\) The value of x is 25.

Solution:

Sum of given terms (\(\sum\)x) = 2.6 min+ 7.2 min+ 3.5 min+ 9.8 min+ 2.5 min =25.6

numbers of terms (n) =5

We have,

Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)

= \(\frac{25.6}{5}\)

= 5.12

Hence, The mean swimming time to the nearest tenth is 5.12 min.

Solution:

Sum of given terms (\(\sum\)x) =

2.7 hr + 8.3 hr + 3.5 hr + 5.1 hr + 4.9 hr = 24.5hr

numbers of terms (n) =5

We have,

Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)

= \(\frac{24.5}{5}\)

= 4.9

Hence, The mean race time is 4.9 hr

Solution:

Now,
Sum of number =m + m+2 + m+4 + m+6 + m+8
= 5m + 20
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, \(\frac{13}{1}\) = \(\frac{5m + 20}{5}\)
or, 65 = 5m + 20
or, 5m = 65 - 20
or, 5m = 45
or, m = \(\frac{45}{5}\)
or, m = 9
\(\therefore\) The value of m is 9.

Solution:

Now,
Sum of a numbers =y + y+3 + y+7 + y+10 + y+13 + y+15
= 6y + 48
We know that,
or, Mean (\(\overline{X}\)) = \(\frac{\sum{x}}{n}\)
or, 18 = \(\frac{6y + 48}{6}\)
or, 108 = 6y + 48
or, 108 - 48 = 6y
or, 6y = 60
or, y = \(\frac{60}{6}\)
or, y = 10
\(\therefore\) The value of y is 10.


Solution:

x f fx
5 2 10
10 6 60
15 10 150
20 5 100
25 2 50
Total 25 370

Now,
or, \(\overline{X}\) = \(\frac{\sum{x}}{n}\)
or, \(\overline{X}\) = \(\frac{370}{25}\)
\(\therefore\) \(\overline{X}\) = 14.8

Solution:

Class Interval frequency Mid-Value F.x
0 - 6 32 3 96
6 - 12 4 9 36
12 - 18 10 15 150
18 - 24 6 21 126
24 - 30 2 27 54
N = 24 \(\sum{fx}\) = 372

We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{372}{24}\)
\(\therefore\)\(\overline{X}\) =15.5

Solution:

Weight No. of students fx
20 4 80
24 6 144
30 8 240
32 5 160
35 3 70
N = 26 \(\sum{fx}\) = 694

We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{694}{26}\)
\(\therefore\) \(\overline{X}\) = 26.69 kg



Solution:

Marks

Students

mid-value

fx

10 - 20

4

15

60

20 - 30

6

25

150

30 - 40

10

35

350

40 - 50

3

45

135

50 - 60

2

55

110

N = 25

\(\sum{fx}\) = 805

We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{372}{25}\)
\(\therefore\) \(\overline{X}\) =14.88

Solution:

Height

No. of plants

fx

58

12

696

60

14

840

62

20

1240

64

13

832

66

8

528

68

5

340

N = 72

\(\sum{fx}\) = 4478

We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{4478}{72}\)
\(\therefore\) \(\overline{X}\) = 62.19 cm

Solution:

Marks

Students

Mid-Value

fx

5 - 10

2

7.5

15

10 - 15

4

12.5

50

15 - 20

8

17.5

140

20 - 25

6

22.5

135

25 - 30

4

27.5

110

N = 24

\(\sum{fx}\) = 450

We know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, \(\overline{X}\) = \(\frac{450}{24}\)
\(\therefore\) \(\overline{X}\) = 18.75

Solution:

class interval x f fx
0 - 8 4 5 20
8 - 16 12 9 108
16 - 24 10 10 200
24 - 32 28 p 28 + p
32 - 40 36 8 288
N = 32 + p \(\sum{fx}\) = 616 + 28p

we know that,
or, \(\overline{X}\) = \(\frac{\sum{fx}}{N}\)
or, 21= \(\frac{616 + 28p}{32 + p}\)
or, 672 + 21p = 616 + 28p
or, 672 - 616 = 28p - 21p
or, 56 = 7p
or, p = \(\frac{56}{7}\)
or, p = 8
\(\therefore\) The required mean is 8.