Question:
A set of data contains 53 observations. The minimum value is 42 and the maximum value is 129. The data are to be organized into a frequency distribution.
a. How many classes would you suggest?
b. What would you suggest as the lower limit of the first class?
Answers
Answered by
2
Answer:
RESULTS:
A. 6 CLASSES
B. 40
Explaining:
A. IF K IS THE NUMBER OF CLASSES AND N IS THE NUMBER OF OBSERVATIONS, THEN FOR CLASS INTERVALS, SELECT THE SMALLEST K SUCH THAT 2 > N.
HERE, N 53. IF K 5, THEN 2 = 25 = 32 WHICH IS LESS THAN N = 53. so. K = 5 CANNOT BE USED. IFK = 6, THEN 2 = 26 = 64 WHICH IS GREATER THAN N = 53. so, NUMBER OF CLASSES WILL BE 6.
B. SEE THE ABOVE ATTACHMENT
Step-by-step explanation:
100% copied answer
Attachments:
Answered by
1
Answer:
a. If k denotes the number of classes and n states the number of observations in the statistical data. Then it could be said that this equation should be satisfied, which is as under:
2k = n
If k = 5, then 2⁵ = 32 < 63
If k - 6,then 2⁶ = 64 > 63
Hence, k should be equated with 6, which gives the number of classes equivalent to 6.
b. Width of the class would be determined as,
= Max value - min value / k
= 129 - 42/6
= 87/6
= 14.5
It could be said that the 14.5 is the multiple of 5, whereas the lowest value is 42. If the lower limit is settled at 43.5, which is also the multiple of 14.5. But 43.5 stays at the greater value relative to 42. Hence, 40 could be taken as the lower limit of the class interval. It is also the multiple of 5 and would also entail 42 in this. Hence, the class interval also would be taken as 40-44.5.
HOPE IT HELPS YOU
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