#### The average age of five students is 15 years. If one student is excluded, then the mean gets reduced by two years. The age of the excluded student is ____.

Given the average age of five students is 15 years. ⇒ The sum of the ages of the five students = 15 × 5 = 75 If one student is excluded, then the mean is reduced by two years. ⇒ The average age of four students = 13 years ⇒ The sum of the ages of the four students = 13 × 4 = 52 The age of excluded student = 75 – 52 =23 Hence, the age of the excluded student is 23 years.

#### A teacher calculated the mean marks of 10 students as 83. Afterwards, he increased the marks of one student by 7. Then the new mean is ____.

Given the the mean marks of 10 students is 83. Total marks = 10 × 83 = 830 Marks increased for one student = 7 New total = 830 + 7 = 837 New mean=837/10 = 83.7 Hence, the new mean is 83.7

#### The data having unique mode is called uni-mode.

The data having unique mode is called uni-mode.

#### The formula to find the mode of the grouped frequency distribution is l + (f_{1} – f_{0})/(2f_{1} – f_{0} – f_{2})×h.

The formula to find the mode of the grouped frequency distribution is (f_{1} – f_{0})/(2f_{1} – f_{0} – f_{2})×h.

#### The data having two modes is called tri-modal.

The data having two modes is called bi-modal.

#### The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for the data.

The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for the data.

#### The mean of first n natural numbers is ____.

Mean of first n natural numbers = 1+2+3+....+n/n =[n(n+1)/2]/n =n(n+1)/2n =n +1/2 Hence, the mean of first n natural numbers is n + 1/2.

#### The formula to find the median of a grouped data is l + [(n/2) – cf]/f ×h.

The formula to find the median of a grouped data is l + [(n/2) – cf]/f ×h.

#### The mean of 0.2, 0.1, 0.4 and 0.5 is ____.

Given numbers are 0.2, 0.1, 0.4 and 0.5 Mean = Sum of the observations/Number of the observations = 0.2 + 0.1 + 0.4 + 0.5 /4 =1.2/4 =0.3 Hence, the mean of 0.2, 0.1, 0.4 and 0.5 is 0.3

#### The mean of five observations is 42. If one of the observations is deleted, then the mean of the remaining observations is 40. The deleted observation is ____.

Given the mean of five observations is 42. ⇒ The sum of the five observations = 5 × 42 = 210 Let the deleted observation be x. ⇒ The sum of the four observations = 210 – x The mean of four observations=210 − x/4 ⇒ 210 – x/4 = 40 (Given) ⇒ 210 – x = 40 × 4 ⇒ − x = 160 – 210 ⇒ − x = − 50 ⇒ x = 50 Hence, the deleted observation is 50.