Question

33. The weights in grams of 50 oranges priced at random from a consignment are as
follows:
131, 113, 82, 75, 204, 81, 84 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90,
115. 110, 98, 106, 99, 107.84, 76, 186, 82, 100, 109, 128, 115, 107, 115, 119.93.187.
139, 129, 130, 68, 195, 123, 125, 111, 92, 86, 70, 126.
From the grouped frequency table by dividing the variable range into intervals of equal
width. each corresponding to 20 grams in such a way that the mid-value of the first class
corresponds to 70 grams.​

Answer 1

Answer:

Upper Limit - Lower Limit = Width

It is given Width is 20.

So, Upper Limit - Lower Limit = 20 --- (1)

(Upper Limit + Lower Limit) / 2 = Mid Value

So, (Upper Limit + Lower Limit) / 2 = 70

So, (Upper Limit + Lower Limit)  = 140 --- (2)

Solving (1) and (2) , we get

Upper Limit = 80

Lower Limit = 60

Frequency distribution table as below based on the consignment  data :-

Class Interval       |        Frequency

60-80                             5

80-100                           13

100-120                           17

120-140                           10

140-160                           1

160-180                           0

180-200                           3

200-220                           1

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Answer 2

Answer:

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