The point estimate of the proportion of students who prefer the new schedule is 0.22.
Definition of Point Estimate
Point estimation is a technique used in statistics to determine a single value that will serve as the "best guess" or "best estimate" of an unidentified population characteristic, known as the point estimate. More precisely, in order to produce a point estimate, we apply a point estimator to the data.
For the confidence interval (0.195, 0.245), point estimate is calculated as follows,
p = (0.195 + 0.245) / 2
p = 0.44 / 2
p = 0.22
Hence, the point estimate of the proportion of students who prefer the new schedule comes out to be 2.2.
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It is 10 times greater because
10x10=100
A20-a18=a1+19d-a1-17d=2d =281-97
so
d=92
Part A:
Consider from x = -5 to x = -4, they are 1 unit apart and the difference of their outputs is given by:
-3 - (-11) = -3 + 11 = 8.
Thus, the value of the output increases by 8 units for each one unit increase in the input.
Part B:
Consider from x = -3 to x = -1, they are 2 units apart and the difference of their outputs is given by:
21 - 5 = 16.
Thus, the value of the output increases by 16 units for each two units increase in the input.
Part C:
Consider from x = 0 to x = 3, they are 3 units apart and the difference of their outputs is given by:
53 - 29 = 24.
Thus, the value of the output increases by 24 units for each three units increase in the input.
Part D:
It can be noticed that the ratio difference in the outputs to the input intervals are equal for all the given input intervals.
i.e 8 / 1 = 16 / 2 = 24 / 3.
Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
