To make a box and whisker plot, first you write down all of the numbers from least to greatest.
0, 1, 3, 4, 7, 8, 10
The median is 4, so that’s the middle line of the plot.
So now we have:
0, 1, 3, [4,] 7, 8, 10
So next we have to find the 1st and 3rd interquartiles..
0, [1,] 3, [4,] 7, [8,] 10
Those are the next 2 points you put on the plot.
Lastly, the upper and lower extremes. These are the highest and lowest numbers in the data.
[0,] 1, 3, 4, 7, 8, [10]
These are the final points on the plot.
To make the box of a box-and-whisker plot, you plot the 3 Medians of the data: 1, 4, and 8, and connect those to make a box that has a line in the middle at 4.
Next, you plot the upper and lower extremes: 0 and 10, by making “whiskers” that connect to the box. So you draw a line from the extremes to the box.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.
The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.
A sample of 34 cups was taken:
a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:
The lower value will be:
= -1.648
You reverse the standardization using the formula ![Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~N%280%3B1%29)
![-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }](https://tex.z-dn.net/?f=-1.648%3D%20%5Cfrac%7BX%5Bbar%5D-8%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B34%7D%20%7D%20%7D)
X[bar]= 7.72ounces
The lower control point will be 7.72 ounces.
The upper value will be:

![1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }](https://tex.z-dn.net/?f=1.648%3D%20%5Cfrac%7BX%5Bbar%5D-8%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B34%7D%20%7D%20%7D)
X[bar]= 8.28ounces
The upper control point will be 8.82 ounces.
b. Now μ= 7.6, considering the control limits of a.
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))
P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242
There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.
b. For this item μ= 8.7, the control limits do not change:
P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)
P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))
P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007
There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.
I hope it helps!
Two angles<span> are said to be supplementary when the sum of the two </span>angles<span> is </span>180<span>°. ... All </span>angles<span> that are either exterior </span>angles<span>, </span>interior angles<span>, </span>alternate angles<span> or corresponding </span>angles<span> are all congruent.</span>
Answer:
the correct graph of this function is a graph of a line that has a y-intercept that is approximately negative $20 million and an x-intercept that is approximately 2.7 years.
Step-by-step explanation: