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.