The billiard balls without replacement.

Let's say there are 10 billiard balls 5 red and 5 blue.
If you pick a blue ball on the first pick you had a 5/10 chance
Now your chances of getting a red are only 5/9.
What happened on the first pick affects what happens on the second (making it dependent)

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3 years ago

Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the

Troyanec [42]

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

$\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3$

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}$

Now, we calculate the sum

$\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\$