According to Sturge's rule, number of classes or bins recommended to construct a frequency distribution is k ≈ 7
Sturge's Rule: There are no hard and fast guidelines for the size of a class interval or bin when building a frequency distribution table. However, Sturge's rule offers advice on how many intervals one can make if one is genuinely unable to choose a class width. Sturge's rule advises that the class interval number be for a set of n observations.
Given,
n = 66
We know that,
According to Sturge's rule, the optimal number of class intervals can be determined by using the equation:
Here, n is equal to 66 and by substituting the value to the equation we get:
k = 7.0444
k ≈ 7
Learn more about Sturge's rule here: brainly.com/question/28184369
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8 is the missing number
Let’s say x is 2,
3(2)+8=14
8(2)-2=14
Answer:
66.48% of full-term babies are between 19 and 21 inches long at birth
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean length of 20.5 inches and a standard deviation of 0.90 inches.
This means that 
What percentage of full-term babies are between 19 and 21 inches long at birth?
The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then
X = 21
has a p-value of 0.7123
X = 19
has a p-value of 0.0475
0.7123 - 0.0475 = 0.6648
0.6648*100% = 66.48%
66.48% of full-term babies are between 19 and 21 inches long at birth
I believe C. Because it provides a more meaningful explanation
