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 + (f1 – f0)/(2f1 – f0 – f2)×h.
The formula to find the mode of the grouped frequency distribution is (f1 – f0)/(2f1 – f0 – f2)×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.
