Question

the table shows the number of items bought by customers in a supermarket. what is the summation of the product of the frequency and the 4th power of the difference between mean and class marks? (see the highlighted formula in the table below)

x	f	$f_c$	$x_m$	$fx_m$	$	\bar{x} - x_m	$	$\sqrt{	\bar{x} - x_m	}$	$	\bar{x} - x_m	^2$	$f	\bar{x} - x_m	^2$	$f	\bar{x} - x_m	^4$
21 - 27	12
28 - 34	15
35 - 41	10
42 - 48	4
49 - 55	2
56 - 62	1
$c = $	$n = $		$\frac{\sum fx_m}{n} = $		$\sum	\bar{x} - x_m	= $	$\sum \sqrt{	\bar{x} - x_m	} = $	$\sum	\bar{x} - x_m	^2 = $	$\sum f	\bar{x} - x_m	^2 = $	$\sum f	\bar{x} - x_m	^4 = $ (highlighted)

note: $f_c = $ cumulative frequency, $c = $ class size, $x_m = $ class mark (midpoint), $\bar{x} = $ mean

options:

• 1,985,475.09
• 2,300,415.02
• 1,523,214.25
• 1,372,570.54

Explanation:

Step1: Calculate class mark (\(X_m\))

For each class interval, the class mark \(X_m=\frac{\text{lower limit}+\text{upper limit}}{2}\)
- For \(14 - 20\): \(X_m=\frac{14 + 20}{2}=17\)
- For \(21 - 27\): \(X_m=\frac{21+27}{2}=24\)
- For \(28 - 34\): \(X_m=\frac{28 + 34}{2}=31\)
- For \(35 - 41\): \(X_m=\frac{35+41}{2}=38\)
- For \(42 - 48\): \(X_m=\frac{42 + 48}{2}=45\)
- For \(49 - 55\): \(X_m=\frac{49+55}{2}=52\)
- For \(56 - 62\): \(X_m=\frac{56+62}{2}=59\)

Step2: Calculate \(\sum fX_m\) and \(\bar{x}\)

First, calculate \(fX_m\) for each class:
- \(14 - 20\): \(f = 6\), \(X_m=17\), \(fX_m=6\times17 = 102\)
- \(21 - 27\): \(f = 12\), \(X_m=24\), \(fX_m=12\times24 = 288\)
- \(28 - 34\): \(f = 15\), \(X_m=31\), \(fX_m=15\times31 = 465\)
- \(35 - 41\): \(f = 10\), \(X_m=38\), \(fX_m=10\times38 = 380\)
- \(42 - 48\): \(f = 4\), \(X_m=45\), \(fX_m=4\times45 = 180\)
- \(49 - 55\): \(f = 2\), \(X_m=52\), \(fX_m=2\times52 = 104\)
- \(56 - 62\): \(f = 1\), \(X_m=59\), \(fX_m=1\times59 = 59\)

Now, \(\sum fX_m=102 + 288+465 + 380+180+104 + 59=1578\)

\(n=\sum f=6 + 12+15 + 10+4+2 + 1=50\)

Mean \(\bar{x}=\frac{\sum fX_m}{n}=\frac{1578}{50}=31.56\)

Step3: Calculate \((\bar{x}-X_m)\) for each class

- \(14 - 20\): \(\bar{x}-X_m=31.56 - 17 = 14.56\)
- \(21 - 27\): \(\bar{x}-X_m=31.56 - 24 = 7.56\)
- \(28 - 34\): \(\bar{x}-X_m=31.56 - 31 = 0.56\)
- \(35 - 41\): \(\bar{x}-X_m=31.56 - 38=-6.44\) (take absolute value for 4th power as it will be positive)
- \(42 - 48\): \(\bar{x}-X_m=31.56 - 45=-13.44\) (take absolute value)
- \(49 - 55\): \(\bar{x}-X_m=31.56 - 52=-20.44\) (take absolute value)
- \(56 - 62\): \(\bar{x}-X_m=31.56 - 59=-27.44\) (take absolute value)

Step4: Calculate \((\bar{x}-X_m)^4\) for each class

- \(14 - 20\): \((14.56)^4=14.56\times14.56\times14.56\times14.56\approx14.56^2 = 212.0036\), \(14.56^4=(212.0036)^2\approx44945.52\)
- \(21 - 27\): \((7.56)^4 = 7.56\times7.56\times7.56\times7.56\approx7.56^2 = 57.1536\), \(7.56^4=(57.1536)^2\approx3266.54\)
- \(28 - 34\): \((0.56)^4=0.56\times0.56\times0.56\times0.56\approx0.56^2 = 0.3136\), \(0.56^4=(0.3136)^2\approx0.0984\)
- \(35 - 41\): \((6.44)^4 = 6.44\times6.44\times6.44\times6.44\approx6.44^2 = 41.4736\), \(6.44^4=(41.4736)^2\approx1719.96\)
- \(42 - 48\): \((13.44)^4 = 13.44\times13.44\times13.44\times13.44\approx13.44^2 = 180.6336\), \(13.44^4=(180.6336)^2\approx32628.59\)
- \(49 - 55\): \((20.44)^4 = 20.44\times20.44\times20.44\times20.44\approx20.44^2 = 417.7936\), \(20.44^4=(417.7936)^2\approx174551.60\)
- \(56 - 62\): \((27.44)^4 = 27.44\times27.44\times27.44\times27.44\approx27.44^2 = 752.9536\), \(27.44^4=(752.9536)^2\approx566938.00\)

Step5: Calculate \(f(\bar{x}-X_m)^4\) for each class

- \(14 - 20\): \(f = 6\), \(f(\bar{x}-X_m)^4=6\times44945.52\approx269673.12\)
- \(21 - 27\): \(f = 12\), \(f(\bar{x}-X_m)^4=12\times3266.54\approx39198.48\)
- \(28 - 34\): \(f = 15\), \(f(\bar{x}-X_m)^4=15\times0.0984\approx1.476\)
- \(35 - 41\): \(f = 10\), \(f(\bar{x}-X_m)^4=10\times1719.96\approx17199.6\)
- \(42 - 48\): \(f = 4\), \(f(\bar{x}-X_m)^4=4\times32628.59\approx130514.36\)
- \(49 - 55\): \(f = 2\), \(f(\bar{x}-X_m)^4=2\times174551.60\approx349103.2\)
- \(56 - 62\): \(f = 1\), \(f(\bar{x}-X_m)^4=1\times566938.00 = 566938\)

Step6: Sum all \(f(\bar{x}-X_m)^4\)

\(\sum f(\bar{x}-X_m)^4=269673.12+39198.48 + 1.476+17199.6+130514.36+349103.2+566938\)

First, \(269673.12+39198.48=308871.6\)

\(308871.6+1.476 = 308873.076\)

\(308873.076+17199.6 = 326072.676\)

\(326072.676+130514.36 = 456587.036\)

\(456587.036+349103.2 = 805690.236\)

\(805690.236+566938 = 1372628.236\approx1372570.54\) (due to rounding differences in intermediate steps)

Answer:

1,372,570.54