### Question 11. Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

## Conc. of SO | ## 0.00-0.04 | ## 0.04-0.08 | ## 0.08-0.12 | ## 0.12-0.16 | ## 0.16-0.20 | ## 0.20-0.24 |

## No of days | ## 4 | ## 8 | ## 9 | ## 2 | ## 4 | ## 3 |

### Find the probability of concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.

**Solution:**

Total number of days = 30

Probability of concentration of SO

^{2}in the internal 0.12 – 0.16 == Favorable Outcome / Total outcome

= 2/30 = 0.06

### Question 12. An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

** **Vehicles per family

## Monthly Income(in Rs) | ## 0 | ## 1 | ## 2 | ## Above 2 |

## Less than 7000 | ## 10 | ## 160 | ## 25 | ## 0 |

## 7000-10000 | ## 0 | ## 305 | ## 27 | ## 2 |

## 10000-13000 | ## 1 | ## 535 | ## 29 | ## 1 |

## 13000-16000 | ## 2 | ## 469 | ## 59 | ## 25 |

## 16000 or more | ## 1 | ## 579 | ## 82 | ## 88 |

### Suppose a family is chosen, find the probability that the family chosen is

### (i) earning Rs 10000 − 13000 per month and owning exactly 2 vehicles.

### (ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.

### (iii) earning less than Rs 7000 per month and does not own any vehicle.

### (iv) earning Rs 13000 − 16000 per month and owning more than 2 vehicles.

### (v) owning not more than 1 vehicle.

### (vi) owning at least one vehicle.

**Solution:**

Total numbers of families selected by the organization to Survey= 2400. -(According to Question)

(i)Let E1 be the event of selecting of family earning ₹(10000 -13000)per month and owning exactly two vehicles.

Numbers of families earning ₹10000 –13000

per month and owning exactly 2 vehicles = 29

Required probability P(E1) = 29/2400

(ii)Let E2 be the event of selecting of family earning ₹16000 ormore per month and owning exactly 1 vehicle.

Number of families earning ₹16000 or

more per month and owning exactly 1 vehicle = 579

Required probability,P(E2) = 579/2400

(iii)Let E3 be the event of selecting of family earning than ₹ 7000 per month anddoesn’t own any vehicle.

Number of families earning but ₹7000 per month and

doesn’t own any vehicle = 10

Required probability, P(E3)= 10/2400 = 1/240

(iv)Let E4 be the event of selecting a family earning ₹(13000 -16000) per month andowning quite 2 vehicles.

Number of families earning ₹13000-16000 per month and

owning quite 2 vehicles = 25

Required probability, P(E4) = 25/2400 = 1/96

(v)Let E5 be the event of selecting a family owning less than 1 vehicle.Number of families owning less than 1 vehicle i.e. the number

of families owning 0 vehicle and 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2069

Required probability, P(E5) = 2062/2400 = 1031/1200

(vi)Let P(E6) is the probability that the family of owning atleast one vehicleP(E6) = (160+305+535+469+579+25+27+29+29+82+0+2+1+25+88)/2400

= 2356/2400 = 589/600

### Question 13. The following table gives the lifetimes of 400 neon lamps :

## Lifetime | ## 300-400 | ## 400-500 | ## 500-600 | ## 600-700 | ## 700-800 | ## 800-900 | ## 900-1000 |

## Number of lambs | ## 14 | ## 56 | ## 60 | ## 86 | ## 74 | ## 62 | ## 48 |

### A bulb is selected in random. Find the probability that the lifetime of the selected bulb is

### (i) less than 400?

### (ii) Between 300 to 800?

### (iii) At least 700hours?

**Solution:**

(i)The probability that the lifetime of the selected bulb is less than 400= Favorable outcomes / Total outcome

= 14/400 = 7/400

(ii)The probability that the lifetime of the selected bulb is between 300 – 800 hours= Favorable outcomes / Total outcome

= (14 +56 +60 +86 +74) / 400

= 29/40

(iii)The probability that the lifetime of the selected bulb is at least 700 hours= Favorable outcomes / Total outcome

= (74 +62+ 48)/400 = 23/50

### Question 14. Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

## Wages(in Rs) | ## 110-130 | ## 130-150 | ## 150-170 | ## 170-190 | ## 190-210 | ## 210-230 | ## 230-250 |

## No. of workers | ## 3 | ## 4 | ## 5 | ## 6 | ## 5 | ## 4 | ## 3 |

### A worker is selected at random. Find the probability that

### (i) Less than Rs150

### (ii) Atleast Rs210

### (iii)More than or equal to Rs150 but less than Rs210.

**Solution:**

(i)The probability that his wages are less than Rs 150 == Favorable outcomes / Total outcome

=(3 + 4) / 30 = 7 / 30

(ii)The probability that his wages are at least Rs 210= Favorable outcomes / Total outcome

= (3 + 4) / 30 = 7 / 30

(iii)The probability that his wages are more than or equal to 150 but less than Rs 200= Favorable outcomes / Total outcome

= (5 + 6 + 5) / 30 = 16 / 30 = 8 / 15