# Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

### Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean, and mode of the data and compare them.

Monthly consumption(in units) | No. of customers |

65-85 | 4 |

85-105 | 5 |

105-125 | 13 |

125-145 | 20 |

145-165 | 14 |

165-185 | 8 |

185-205 | 4 |

**Solution:**

Total number of consumer n = 68

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l= 125, n = 68, C_{f }= 22, f = 20, h = 20Now we find the median:

Median =

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f

_{1 }= 20, f_{0 }= 13, f_{2 }= 14 & h = 20Mode

=

On substituting the values in the given formula, we getMode =

= 125 + 140/13

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Intervalf_{i}x_{i}d_{i }= x_{i }– au_{i }= d_{i}/hf_{i}u_{i}65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12 Sum f_{i }= 68Sum f_{i}u_{i }= 7= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

### Question 2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Class Interval | Frequency |

0-10 | 5 |

10-20 | x |

20-30 | 20 |

30-40 | 15 |

40-50 | y |

50-60 | 5 |

Total | 60 |

**Solution:**

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class,

l= 20,C

_{f}= 5 + x,f = 20 & h = 10

Now we find the median:

Median =

On substituting the values in the given formula, we get28.5 =

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

60 = 5 + 20 + 15 + 5 + x + y

On substituting the values ofx, we will find the value of y60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

So the value of a is 8 and y is 7

### Question 3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Age (in years) | Number of policy holder |

Below 20 | 2 |

Below 25 | 6 |

Below 30 | 24 |

Below 35 | 45 |

Below 40 | 78 |

Below 45 | 89 |

Below 50 | 92 |

Below 55 | 98 |

Below 60 | 100 |

**Solution: **

According to the given question the table is

Class intervalFrequencyCumulative frequency15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100 Given data: n = 100 and n/2 = 50

Median class = 35 – 45

Then,

l= 35, c_{f}= 45, f = 33 & h = 5Now we find the median:

Median =

On substituting the values in the given formula, we getMedian =

= 35 + 5(5/33)

= 35.75

Hence, the median age is 35.75 years.

### Question 4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Length (in mm) | Number of leaves |

118-126 | 3 |

127-135 | 5 |

136-144 | 9 |

145-153 | 12 |

154-162 | 5 |

163-171 | 4 |

172-180 | 2 |

### Find the median length of leaves.

**Solution:**

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

Class IntervalFrequencyCumulative frequency117.5-126.5 3 3 126.5-135.5 5 8 135.5-144.5 9 17 144.5-153.5 12 29 153.5-162.5 5 34 162.5-171.5 4 38 171.5-180.5 2 40 From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l= 144.5,cf = 17, f = 12 & h = 9

Now we find the median:

Median =

On substituting the values in the given formula, we getMedian =

= 144.5 + 9/4

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

### Question 5. The following table gives the distribution of a life time of 400 neon lamps.

Lifetime (in hours) | Number of lamps |

1500-2000 | 14 |

2000-2500 | 56 |

2500-3000 | 60 |

3000-3500 | 86 |

3500-4000 | 74 |

4000-4500 | 62 |

4500-5000 | 48 |

### Find the median lifetime of a lamp.

**Solution:**

According to the question

Class IntervalFrequencyCumulative1500-2000 14 14 2000-2500 56 70 2500-3000 60 130 3000-3500 86 216 3500-4000 74 290 4000-4500 62 352 4500-5000 48 400 n = 400 and n/2 = 200

Median class = 3000 – 3500

l= 3000, C_{f }= 130,f = 86 & h = 500

Now we find the median:

Median =

On substituting the values in the given formula, we getMedian =

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

### Question 6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |

Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |

### Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

**Solution:**

According to the question

Class IntervalFrequencyCumulative Frequency1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100 n = 100 and n/2 = 50

Median class = 7 – 10

Therefore,

l= 7, C_{f}= 36, f = 40 & h = 3Now we find the median:

Median =

On substituting the values in the given formula, we getMedian =

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

Where,

l= 7, f_{1}= 40, f_{0}= 30, f_{2}= 16 & h = 3Mode =

On substituting the values in the given formula, we getMode =

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Intervalf_{i}x_{i}f_{i}x_{i}1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70 Sum f_{i}= 100Sum f_{i}x_{i}= 825Mean =

= 825/100 = 8.25

Hence, the mean is 8.25

### Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Weight(in kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |

Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |

**Solution:**

According to the question

Class IntervalFrequencyCumulative frequency40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30 n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, C

_{f}= 13, f = 6 & h = 5Now we find the median:

Median =

On substituting the values in the given formula, we getMedian =

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67