The class mid-points in a frequency distribution table are $125, 132, 139, 146$ and $153$ . The class boundaries of the last class are

135.5 - 142.5

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136 - 143

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149.5 - 156.5

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150 - 157

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Solution

The correct option is C $149.5 - 156.5$
Lower limit $=$ Mean of the last two mid values of the classes $= \frac{146 + 153}{2} = 149.5$
Difference between Mid value $-$ Lower limit $=$ upper limit $-$ mid value
Hence, upper limit $= 2 \times 153 - 149.5 = 156.5$

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The class mid-points in a frequency distribution table are 125,132,139,146 and 153. The class boundaries of the last class are

The correct option is C 149.5−156.5Lower limit = Mean of the last two mid values of the classes =146+1532=149.5Difference between Mid value − Lower limit = upper limit − mid valueHence, upper limit =2×153−149.5=156.5

The class mid-points in a frequency distribution table are $125, 132, 139, 146$ and $153$ . The class boundaries of the last class are

The correct option is C $149.5 - 156.5$
Lower limit $=$ Mean of the last two mid values of the classes $= \frac{146 + 153}{2} = 149.5$
Difference between Mid value $-$ Lower limit $=$ upper limit $-$ mid value
Hence, upper limit $= 2 \times 153 - 149.5 = 156.5$