In: Math
For the following data, approximate the sample mean, sample variance and sample standard deviation weekly grocery bill.
Bill (in dollars) Frequency
135-139 11
140-144 20
145-149 19
150-154 7
155-159 5
Solution:
Given that,
Class | Frequancy ( f ) | Mid value ( x ) | d = x -A /h | f*d | fd2 |
135-139 | 11 | 137 | -2 | -22 | 44 |
140 - 144 | 20 | 142 | -1 | -20 | 20 |
145 -149 | 19 | 147 | 0 | 0 | 0 |
150 - 154 | 7 | 152 | 1 | 7 | 7 |
155 - 159 | 5 | 157 | 2 | 10 | 20 |
n = 62 | ![]() |
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a ) The sample mean is
Mean = (A + (
fd / n) * h)
= (147 +( - 25 / 62 ) * 5
= 147 + - 0.4032 * 5
= 147 + - 2.0161
= 144.9839
b ) The sample variance S2
S2 = ( fd2 ) - ((
fd )2 / n ) / 1 -n ) *
h2
= ( 91 ( (- 25 )2 / 62 ) / 61 * 52
= ( 91 -10.0806 / 61 ) * 25
= (80.9194 / 61 )* 25
= 1.3262 *25
= 33.1637
The sample variance S2 is 33.1637
c ) The sample standard is S
S =
(
fd2 ) - ((
fd )2 / n ) / 1 -n ) *
h2
=
( 91 ( (- 25 )2 / 62 ) / 61 * 52
=
( 91 -10.0806 / 61 ) * 25
=
(80.9194 / 61 )* 25
=
1.3262 *25
=
33.1637
= 5.7588
The sample standard is = 5.7588