Answer to Question #142075 in Statistics and Probability for chilekwa

Question

In one month, a student recorded the lengths, to the nearest minute,
of each of the lectures she attended. The table below shows her data
and the calculations she made before drawing a histogram to illustrate
these data.
Calculate
Length of lecture (minutes) 50 ≤ x < 53 53 ≤ x < 55 55 ≤ x < 59 59 ≤ x < 67
Number of lectures a b 30 c
Frequency density 5 13 7.5 1.5
(i) The value of a, of b and of c.
(ii) The number of lectures attended in a month.
(iii) Using the assumed mean of A = 57.4, calculate the mean length
of lectures.
(iv) Calculate the Median and the Variance of the distribution.

Answer 1

(i) a = 30/7.5*5 = 20

b = 30/7.5*13 = 52

c = 30/7.5*1.5 = 6

(ii) 20+52+30+6 = 108

(iii) 53 min 53 sec = 53.9

55 55 = 55.9

59 59 = 60.0

L1 = (53.9+50)/2 =52.0

L2 = (53.9+55.9)/2 = 54.9

L3 = (60+55.9)/2 = 58.0

L4 = (60+67)/2 = 63.7

mean length of lectures = (52.0*5+54.9*13+58.0*7.5+63.7*1.5)/(5 +13+ 7.5+ 1.5) = 55.7 (lower than assumed)

(iv) Median value is ((108+1)/2)^th value = 54.5^th - it is from the second range

median value = 54.9

$\mu$ - mean

$\sigma$² = $\sum$ [X²]/N-$\mu$² = (52.0² *20+54.9² *52+58.0²*30+63.7²*6)/108-54.9² = 97.8