Ex.14.3 Q5 Statistics Solution - NCERT Maths Class 10

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Question

The following table gives the distribution of the life time of \(400\) neon lamps:

Life time (in hours) Number of lamps
\(1500 - 2000\) \(14\)
\(2000 - 2500\) \(56\)
\(2500 - 3000\) \(60\)
\(3000 - 3500\) \(86\)
\(3500 - 4000\) \(74\)
\(4000 - 4500\) \(62\)
\(4500 - 5000\) \(48\)

Find the median life time of a lamp.

   

Text Solution

  

What is known?

The life time of \(400\) neon lamps.

What is unknown?

The median life time of a lamp.

Reasoning:

Median Class is the class having Cumulativefrequency \((cf)\) just greater than \(\frac n{2}\)

Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)

Class size,\(h\)

Number of observations,\(n\)

Lower limit of median class,\(l\)

Frequency of median class,\(f\)

Cumulative frequency of class preceding median class,\(cf\)

Steps:

Life time (in hours) Number of lamps \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) Cumulative frequency
\(1500 - 2000\) \(14\) \(14\)
\(2000 - 2500\) \(56\) \(14 +56 =70\)
\(2500 - 3000\) \(60\) \(70 +60 =130\)
\(3000 - 3500\) \(86\) \(130 +86 =216\)
\(3500 - 4000\) \(74\) \(216 + 74 = 290\)
\(4000 - 4500\) \(62\) \(290 + 62 = 352\)
\(4500 - 5000\) \(48\) \(352 + 48 = 400\)
Total (n) = 400

From the table, it can be observed that

\(n = 400{\rm{    }} \Rightarrow \frac{n}{2} = 200\)

Cumulative frequency \((cf)\) just greater than \(200 \) is \(216,\)belonging to class \(3000 – 3500.\)

Therefore, median class \(=3000 – 3500\)

Class size (\(h\)) \(=500\).

Lower limit (\(l\)) of median class \(=3000\).

Frequency (\(f\)) of median class \(=86\).

Cumulative frequency (\(cf\)) of class preceding median class \(=130\).

\[\begin{align}\begin{aligned} \text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=3000+\left(\frac{200-130}{86}\right) \times 500 \\ &=3000+\frac{70 \times 500}{86} \\ &=3406.976 \end{aligned}\end{align}\]

Therefore, median life time of lamps is \(3406.98\) hours.

  
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