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.

####
Find the mean of the following data.

x | 5 | 10 | 15 | 20 | 25 |

f | 2 | 6 | 10 | 5 | 2 |

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

####
Find the mean of the following continuous data.

Class Interval | 0 -6 | 6 -12 | 12 - 18 | 18 - 24 | 24 - 30 |

Frequency | 2 | 4 | 10 | 6 | 2 |

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

####
Find the mean of the following data.

Weight (in kg) | 20 | 24 | 30 | 32 | 35 |

No. of students | 4 | 6 | 8 | 5 | 2 |

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

####
Find the mean of the following continuous data.

marks obtained |
10 -20 |
20 - 30 |
30 - 40 |
40 - 50 |
50 - 60 |

No. of students |
4 |
6 |
10 |
3 |
2 |

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

####
Find the mean of the following data.

Height (in cm) |
58 |
60 |
62 |
64 |
66 |
68 |

No. of plants |
12 |
14 |
20 |
13 |
8 |
5 |

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

####
Find the mean of the following continuous data.

Marks |
5 - 10 |
10 - 15 |
15 - 20 |
20 - 25 |
25 - 30 |

No. of students |
2 |
4 |
8 |
6 |
4 |

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

####
The mean of the given data is 21. Find the value of p.

class interval | 0 -8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |

frequency | 5 | 9 | 10 | p | 8 |

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.