Question

5 numbers have a mean of 4, a range of 6, mode of 2 and median of 3. What could the numbers be? ​

Answer 1

$\large \sf \star \: \underline{ \underline{ \bold{solution : }}}$

Given :

• Mean = 4
• Mode = 2
• Median = 3
• Range = 6

$\\ \large \tt \mapsto \: mean = \frac{a + b + c + d + e}{5} = 4 \\ \\ \\ \large \implies \tt \: a + b + c + d + e = 5 \times 4\\ \\ \large \implies \tt \: a + b + c + d + e = 20$

Mode = most frequently occurring value

= 2

Let,

• a = 2
• b = 2

$\\ \boxed{ \rm\leadsto \: range = highest \: \: observation - lowest\: \: observation}$

• Highest observation = x

Range = 6

$\\ \large \tt \implies \: 6 = x - 2 \\ \\ \large \tt \implies \: x = 6 + 2 \\ \\ \large \tt \therefore \: x = 8$

$\\ \large \tt \: median = \frac{n + 1}{2} \: term \\ \\ \\ \large \implies \tt \: median = \frac{6}{2} = 3$

3rd term = 3

Now,

• a = 2
• b = 2
• c = 3
• d = 8
• e = ?

$\\ \large \implies \tt \: a + b + c + d + e = 20\\ \\ \large \implies \tt \: 2 + 2 + 3 + 8+ e = 20 \\ \\ \large \tt \implies \: 15 + e = 20 \\ \\ \large \tt \therefore \: e = 5$

So, the numbers could be :

2,2,3,8,5

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$\\ \large \sf \star \: \underline{ \underline{ \bold{formulas : }}}$

$\\ \sf \: mean = \frac{sum \: \: of \: \: all \: \: observations}{number \: \: of \: \: observations} \\ \\ \\ \sf median = \left( \frac{n}{2} + 1\right )^{th} \: \: term \\ \\ \\ \sf \: range = highest \: \: obs. - lowest \: \: obs. \\ \\ \sf \: mode = most \: \: frequently \: \: occuring$