Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Answer:
12
Step-by-step explanation:
4y - 2x
= 4(7) - 2(8)
= 28 - 16
= 12
Answer:
The probability that a part was manufactured on machine A is 0.3
Step-by-step explanation:
Consider the provided information.
It is given that Half of a set of parts are manufactured by machine A and half by machine B.
P(A)=0.5
Let d represents the probability that part is defective.
Ten percent of all the parts are defective.
P(d) = 0.10
Six percent of the parts manufactured on machine A are defective.
P(d|A)=0.06
Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective
:



Hence, the probability that a part was manufactured on machine A is 0.3
Answer:
Step-by-step explanation:
x² - 12x + a = (x + b)²
x² - 12x + a = x² + 2xb + b²
-12 = 2b and a = b²
b = -6
a = b² = 36