Statistics

Correct option: 2 — 2-3

Correct option: 3 — 5

Correct option: 1 — only mean

Correct option: 2 — p - Median

q - Mean

r - Mode

Correct option: 1 — 5.10 marks

Correct option: 3 — Mode

Correct option: 4 — 18

Correct option: 2 — 72.5

• Writes that the company should fix the size of the oil bottle based on the mode. [0.5]

• Writes that the mode gives the information about the size that is sold most often. [0.5]

• Concludes that 25 comes the maximum number of times hence it is the mode of the data. [1]

• States that Farhan can obtain the median of the data by intersection of less than and more than ogives. [1]

• Arranges the data in ascending order as: 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5 [0.5]

• States that n = 13, therefore, median will be the (n+1)/2 th observation, which is 7th observation, that is 4. [0.5]

• Rewrites the data of channel 1 as:

Class Interval: 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | Total

fᵢ: 10000 | 5000 | 6500 | 1500 | 700 | 1300 | n = 25000

xᵢ: 1 | 3 | 5 | 7 | 9 | 11 |

fᵢxᵢ: 10000 | 15000 | 32500 | 10500 | 6300 | 14300 | 88600 [1.5]

• Writes that, on an average, a viewer watches channel 1 for 88600/25000 = 3.54 hours. [0.5]

• Writes that Sunil's statement is not correct. [0.5]

• Provides a valid example where the mean score of some students is 70 marks, half of them scored 85 marks or above, but the rest have not scored 55 marks or lower. For example, considers 4 students in a class with scores, 89, 85, 56 and 50 and shows that the condition specified by Sunil's statement is not satisfied. [1.5]

• Arranges the given data in ascending order as 11, 14, 14, 14, 16, 17, 18, 18, 18, 19, 19, 20, 23, 24, 25. [0.5]

• States that the number of scores is 15 (odd), and hence, the (15+1)/2 th or 8th observation will be the median. [1]

• States that the median of given data will be 18. Hence, Rajat's score is 18. [0.5]

• Creates cumulative frequency distribution for the given data as:

Number of bags sold: 0-5 | 5-10 | 10-15 | 15-20 | 20-25

Number of Days: x | 4 | 15 | 5 | 8

Cumulative Frequency: x | x + 4 | x + 19 | x + 24 | x + 32 [1]

• Writes that since 12 lies in the range 10-15, 10-15 is the median class. Applies the median formula for the grouped data and solves for x as: 12 = 10 + 5[(x+32)/2 − (x+4)] / 15 ⇒ x = 12 [1]

• Finds the number of days remaining as 30 − 10 − 10 = 10. [0.5]

• Finds the sum of the temperatures of all the 30 days as 34 °C × 30 = 1020 °C. [0.5]

• Finds the sum of the temperatures of first 10 days as 30 °C × 10 = 300 °C. [0.5]

• Finds the sum of the temperatures of next 10 days as 35 °C × 10 = 350 °C. [0.5]

• Finds the sum of the temperatures of last 10 days as 1020 °C − 300 °C − 350 °C = 370 °C. [0.5]

• Finds the mean temperature of the last 10 days as 370/10 = 37 °C. [0.5]

• Completes the frequency distribution table as:

Rainfall (in mm): 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | Total

Frequency (fᵢ): 2 | 6 | x | 7 | 4 | 19 + x

Class mark (xᵢ): 5 | 15 | 25 | 35 | 45 |

fᵢxᵢ: 10 | 90 | 25x | 245 | 180 | 525 + 25x [1.5]

• Writes the equation for mean as: (525 + 25x) / (19 + x) = 27 [0.5]

• Solves the above equation to find the value of x as 6. Hence, writes that the rainfall ranged between 20-30 mm for 6 days. [1]

• i) Calculates the current average of the number of cars crossing traffic light as (15+9+16+16+14)/5 = 70/5 = 14 cars per traffic light. [0.5]

• Calculates actual average number of cars crossing traffic light as 14 − 3 = 11 cars. [0.5]

• Calculate the actual sum of number of all the cars passing through each traffic lights of the city as 11 × 5 = 55 cars. [0.5]

• Calculate the actual number of cars crossing TL3 as 16 − (70 − 5) = 16 − 15 = 1 car. [0.5]

• ii) Finds the mode of the data which is 16 cars. Hence, states that according to the data, traffic light TL4 was the busiest in that minute. [1]

• Finds the total number of balls bowled by the bowler so far as 145 × 25 = 3625. [1]

• Finds the total number of balls bowled after the latest match as 3625 + 25 = 3650 and total number of wickets taken after the latest match as 145 + 5 = 150. [1]

• Finds the new strike rate of the bowler as 3650/150 = 24.33. [1]

• Creates a frequency table for the given data. It may look as follows:

Number of T.V. sold: 1-3 | 3-5 | 5-7 | 7-9 | 9-11

Number of Days: 5 | 7 | 6 | 6 | 7 [1]

• Uses the given frequency table to create a histogram as follows: bars over 1-3, 3-5, 5-7, 7-9, 9-11 with heights 5, 7, 6, 6, 7 (Number of Days) matching the frequency table above. [2]

• (Note: Here, Step Deviation method is used. Give full marks for any other method used and solved correctly.) Creates following table to calculate the mean of all the grades using Step Deviation method:

Height (in cm): 120-130 | 130-140 | 140-150 | 150-160 | 160-170 | 170-180 | Total

Number of students: 38 | 40 | 46 | 23 | 25 | 8 | 180 [1]

• Takes class size h as 10 and considers a value for assumed mean (a). For example, a = 145 cm. [0.5]

• Creates following table to calculate mean:

Class Interval: 120-130 | 130-140 | 140-150 | 150-160 | 160-170 | 170-180 | Total

Frequency (fᵢ): 38 | 40 | 46 | 23 | 25 | 8 | 180

Class Mark (xᵢ): 125 | 135 | 145 | 155 | 165 | 175 |

dᵢ = xᵢ − a: −20 | −10 | 0 | 10 | 20 | 30 |

uᵢ = dᵢ/h: −2 | −1 | 0 | 1 | 2 | 3 |

fᵢuᵢ: −76 | −40 | 0 | 23 | 50 | 24 | −19 [2]

• Calculates mean using the Step Deviation formula as: x̄ = 145 + 10×(−19)/180 = 145 − 190/180 = 145 − 1.06 = 143.94 cm. (Note: Award 0.5 marks if the student has only written the formula correctly.) [1.5]

• Takes assumed mean, a = 5 plants and creates following table:

Class Interval: 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | Total

Frequency (fᵢ): 4 | 3 | 2 | 5 | 6 | 20

Class Mark (xᵢ): 1 | 3 | 5 | 7 | 9 |

dᵢ = xᵢ − a: −4 | −2 | 0 | 2 | 4 |

fᵢdᵢ: −16 | −6 | 0 | 10 | 24 | 12 [1]

• Calculates mean using assumed mean formula, x̄ = a + Σfᵢdᵢ/Σfᵢ = 5 + 12/20 = 5.6 potted plants. [1]

• Takes assumed mean, a = 6 plants and creates following table:

Class Interval: 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | Total

Frequency (fᵢ): 4 | 3 | 2 | 5 | 6 | 20

Class Mark (xᵢ): 1 | 3 | 5 | 7 | 9 |

dᵢ = xᵢ − a: −5 | −3 | −1 | 1 | 3 |

fᵢdᵢ: −20 | −9 | −2 | 5 | 18 | −8 [1]

• Calculates mean using assumed mean formula, x̄ = a + Σfᵢdᵢ/Σfᵢ = 6 + (−8)/20 = 6 − 2/5 = 5.6 potted plants. [1]

• Concludes that their result will be same. [1]

• Interprets the data and draws correct histogram as given: bars over 0-10, 10-20, 20-30, 30-40, 40-50 with heights 5, 10, 8, 15, 12 (Number of Matches) matching the frequency table above. [2]

• (Note: Here, Step Deviation method is used. Give full marks for any other method used and solved correctly.) Takes class size h as 10 and considers a value for assumed mean (a). For example, a = 25 overs.

Class Interval: 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | Total

Frequency (fᵢ): 5 | 10 | 8 | 15 | 12 | 50

Class Mark (xᵢ): 5 | 15 | 25 | 35 | 45 |

dᵢ = xᵢ − a: −20 | −10 | 0 | 10 | 20 |

uᵢ = dᵢ/h: −2 | −1 | 0 | 1 | 2 |

fᵢuᵢ: −10 | −10 | 0 | 15 | 24 | 19 [1]

• Calculates mean, x̄ = a + h×Σfᵢuᵢ/Σfᵢ = 25 + 10×19/50 = 25 + 19/5 = 25 + 3.8 = 28.8 overs. [1]

• States that most innings were played for 30-40 overs. [0.5]

• Writes that 30-40 overs is the modal class and hence, the mode will definitely be found within this range. [0.5]