Answer to Question #254739 in Statistics and Probability for berna

Question

(a)The table below show Business statistics I examination marks for a class of one hundred students in Mount Kenya university.

Marks

10-19

20-29

30-39

40-49

50-59

60-69

Number of students

4

8

a

22

48

b

The mean score of the students was 46.5

         (i) Show that the values of a and b are 12 and 6 respectively.                       (5marks)

         (ii) Estimate the median for the above sample.                                               (3marks)

        (iii) Determine the variance and the standard deviation                                  (4 marks)     

        (iv) Given that 32% of the students failed the examination, what was the cut off points.

(3 marks)

(v)Estimate the mode for this data.                                                                (3marks)

Answer 1

i)

mean:

$\mu=\dfrac{\sum x_i}{n}=\dfrac{14.5\cdot4+24.5\cdot8+34.5a+44.5\cdot22+54.5\cdot48+64.5b}{4+8+a+22+48+b}=46.5$

$n=4+8+a+22+48+b=100$

Then,

$a+b=18$

$\implies3849+34.5a+64.5b=4650$

$\implies34.5(18-b)+64.5b=801$

$\implies30b=180$

$b=180/30=6,\ a=18-6=12$

$\boxed{a=12}\ \ \boxed{b=6}$

ii)

$median=l + [(n/2−c)/f] × h$

where,

• l = lower limit of median class
• n = total number of observations
• c = cumulative frequency of the preceding class
• f = frequency of median class
• h = class size

cumulative frequency:

10-19: 4

20-29: 12

30-39: 24

40-49: 46

50-59: 94

60-69: 100

median class: 50-59

$median=50+[(100/2-46)/94]\cdot9=50.38$

iii)

variance:

$\sigma^2=\dfrac{\sum f_iM_i^2-n\mu^2}{n-1}$

where f_i is frequency of i-th range,

M_i is midpoint of i-th range.

$\sigma^2=\dfrac{4\cdot14.5^2+8\cdot24.5^2+12\cdot34.5^2+22\cdot44.5^2+48\cdot54.5^2+6\cdot64.5^2-100\cdot46.5^2}{99}\\\ \\=\dfrac{231025-216225}{99}=149.49$

standard deviation:

$\sigma=\sqrt{149.49}=12.23$

iv)

32% percentile:

$P_{32}=l+w(32-c)/f$

where l is the lower boundary of the class containing P₃₂,

w is the class interval size of the class containing P₃₂,

f is the frequency of the class containing P₃₂,

c is the cumulative frequency of the class preceding the class containing P₃₂

class containing P₃₂: 40-49

Then:

$P_{32}=40+9(32-24)/22=43.27$

v)

$mode=l+\dfrac{f_m-f_1}{(f_m-f_1)(f_m-f_)}\cdot h$

where

l is the lower limit of the modal class,

f_m is the frequency of the modal class,

f₁ is the frequency of class preceding the modal class,

f₂ is the frequency of class succeeding the modal class,

h is the width of the modal class.

modal class (a class that has the highest frequency): 50-59

$mode=50+\dfrac{48-22}{48-22+48-6}\cdot9=53.44$