Question

the table shows the number of items bought by customers in a supermarket. after the completion of all the data in the table, what is the computed mean?

x	f	$f_c$	$x_m$	$fx_m$	$	\bar{x}-x_m	$	$f	\bar{x}-x_m	$	$	\bar{x}-x_m	^2$	$f	\bar{x}-x_m	^2$	$f	\bar{x}-x_m	^4$
14 - 20	6
21 - 27	12
28 - 34	15
35 - 41	10
42 - 48	4
49 - 55	2
56 - 62	1
$c = $	$n = $		$\sigma fx_m$	$\sigma f	\bar{x}-x_m	$	$\sigma f	\bar{x}-x_m	^2$	$\sigma f	\bar{x}-x_m	^4$

note: $f_c = $ cumulative frequency, $c = $ class size, $x_m = $ class mark (midpoint), $\bar{x} = $ mean
\bigcirc 31.56
\bigcirc 24.26
\bigcirc 45.74
\bigcirc 29.17

Explanation:

Step1: Find class midpoints (\(X_m\))

For each class interval, the midpoint \(X_m\) is calculated as \(\frac{\text{lower limit} + \text{upper limit}}{2}\).

• For \(14 - 20\): \(X_{m1}=\frac{14 + 20}{2}=17\)
• For \(21 - 27\): \(X_{m2}=\frac{21 + 27}{2}=24\)
• For \(28 - 34\): \(X_{m3}=\frac{28 + 34}{2}=31\)
• For \(35 - 41\): \(X_{m4}=\frac{35 + 41}{2}=38\)
• For \(42 - 48\): \(X_{m5}=\frac{42 + 48}{2}=45\)
• For \(49 - 55\): \(X_{m6}=\frac{49 + 55}{2}=52\)
• For \(56 - 62\): \(X_{m7}=\frac{56 + 62}{2}=59\)

Step2: Calculate \(fX_m\) for each class

• \(14 - 20\): \(f_1X_{m1}=6\times17 = 102\)
• \(21 - 27\): \(f_2X_{m2}=12\times24 = 288\)
• \(28 - 34\): \(f_3X_{m3}=15\times31 = 465\)
• \(35 - 41\): \(f_4X_{m4}=10\times38 = 380\)
• \(42 - 48\): \(f_5X_{m5}=4\times45 = 180\)
• \(49 - 55\): \(f_6X_{m6}=2\times52 = 104\)
• \(56 - 62\): \(f_7X_{m7}=1\times59 = 59\)

Step3: Find total frequency (\(n\)) and total \(fX_m\)

• Total frequency \(n=\sum f=6 + 12+15 + 10+4 + 2+1=50\)
• Total \(fX_m=\sum fX_m=102 + 288+465 + 380+180 + 104+59 = 1578\)

Step4: Compute the mean (\(\bar{X}\))

The formula for the mean of grouped data is \(\bar{X}=\frac{\sum fX_m}{n}\)
Substitute the values: \(\bar{X}=\frac{1578}{50}=31.56\)

Answer:

31.56