Problem sheet – 1

- A contractor employs three types of workers : male, female and children. He pays Rs.40, Rs.30 and Rs.25 per day to a male, female and child worker respectively. Suppose he employs 20 males, 15 females and 10 children, what is the average wage per day paid by the contractor? Would it make any difference in the answer if the number of males, females and children employed are equal.
- A company manufactures performance equipment for cars used in various types of racing. It has gathered the following information on the number of models of engines in different size categories used in the racing market it serves:

Average Wage=(40*20+30*15+25*10)/(20+15+10)= Rs 33.33

Yes it will make a difference, if the number of males, females and children employed are equal. Average wage in that case will be simply average of 40, 30 & 25, which will be Rs 31.66

Class Frequency

(Engine size in cubic inches) (# of models)

101 – 150 1

151 – 200 7

201 – 250 7

251 – 300 8

301 – 350 17

351 – 400 16

401 – 450 15

451 – 500 7

Construct an appropriate graph that will help you answer these questions:

- Seventy percent of the engine models available are larger than about what size?
- What was the approximate middle value in the original data set?
- If the company has designed a fuel-injection system that can be used on racing engines up to 400 cubic inches, about what percentage of the engine models available will not be able to use the system?
- In flying a distance of 1,200 miles between two airports a cargo aircraft encountered strong headwinds and made the trip at a ground speed of 300 miles per hour. The 1,200 miles return flight was made at a ground speed of 400 miles per hour. Calculate the appropriate mean of ground speeds.
- Public transportation and the automobile are two methods an employee can use to get to work each day. Samples of times recorded for each method are shown.

Interval Engine size in cubic inches |
Frequency |
% frequency |
< % cf |
> % cf |

101 – 150 |
1 |
1.28% |
1.28% |
100.00% |

151 – 200 |
7 |
8.97% |
10.26% |
98.72% |

201 – 250 |
7 |
8.97% |
19.23% |
89.74% |

251 – 300 |
8 |
10.26% |
29.49% |
80.77% |

301 – 350 |
17 |
21.79% |
51.28% |
70.51% |

351 – 400 |
16 |
20.51% |
71.79% |
48.72% |

401 – 450 |
15 |
19.23% |
91.03% |
28.21% |

451 – 500 |
7 |
8.97% |
100.00% |
8.97% |

78 |

It will be Harmonic Mean of two speeds = 2/(1/300+1/400)=342.85 mph

Times are in minutes.

Public Transportation: 28 29 32 37 33 25 29 32 41 34

Automobile: 29 31 33 32 34 30 31 32 35 33

- Compute the mean and standard deviation for each method.
- On the basis of your results from (a), which method of transportation should be preferred? Explain.
- Develop a box plot for each method. Does a comparison of the box plots support your conclusion in (c)?

- Public Transportation
- Automobile should be preferred as a means of transport as even though mean is same in both cases the Standard Deviation is smaller for automobile transport. This means probability of employee taking much longer/shorter time than 32 minutes is least. In other words probability of employee arriving within 32 minutes by using automobile is higher than by using public transport.

Time in minutes x |
x
ç _{i
}- x ç |
x
ç _{i
}- x ç ^{2} |

25 |
7 |
49 |

28 |
4 |
16 |

29 |
3 |
9 |

29 |
3 |
9 |

32 |
0 |
0 |

32 |
0 |
0 |

33 |
1 |
1 |

34 |
2 |
4 |

37 |
5 |
25 |

41 |
9 |
81 |

å x |
å = 194 |

Mean = å x_{i}
/N = 320/10 = 32

SD = Ö [194/10] = 4.40

Automobile

Time in minutes x |
x
ç _{i
}- x ç |
x
ç _{i
}- x ç ^{2} |

29 |
3 |
9 |

30 |
2 |
4 |

31 |
1 |
1 |

31 |
1 |
1 |

32 |
0 |
0 |

32 |
0 |
0 |

33 |
1 |
1 |

33 |
1 |
1 |

34 |
2 |
4 |

35 |
3 |
9 |

å x |
å = 30 |

Mean = å x_{i}
/N = 320/10 = 32

SD = Ö [30/10] = 1.732

c) Public Transportation

Q1=29

Median=Q2=32

Q3=34

Automobile

Q1=31

Median=Q2=32

Q3=33

- The price per unit of a commodity has changed as below:

Jan’ 1985 Rs. 25

Jan’ 1990 Rs. 30

Jan’ 1995 Rs. 35

Jan’ 2000 Rs. 42

Calculate the average rate of increase in prices during the period 1985 – 2000.

Method 1. Average rate of increase = (42/25)

^{1/15}=1.0352 which is 3.52% year per year

Method 2.

Jan’ 1985 |
Rs. 25 |
Ratio of increase |

Jan’ 1990 |
Rs. 30 |
1.20 |

Jan’ 1995 |
Rs. 35 |
1.167 |

Jan’ 2000 |
Rs. 42 |
1.20 |

GM of increase ratios = (1.2*1.167*1.2)

^{1/3}=1.188 which is 18.8% increase per 5 year or 18.8/5=3.6% increase per annum

Correct way for % increase per annum would be to take 5

^{th}root of 18.8% increase in 5 years which will be 1.188^{1/5 }= 1.0352 i.e. 3.52%

- A national door-to-door selling organization released the following information on earnings of 3,850 college students employed last summer in a special selling campaign: maximum earnings $2,830.50, minimum earnings $263.40, mean $1,730.50, median $1.564.40, standard deviation $215.25.
- In applying the Chebyshev inequality, what can be said about the proportion of students earning less than $1,300.00 or more than $2,161.00?
- What can be said about the proportion of students earning between $869.50 and $2,591.50?