Two ballpoint pens are selected at random from a box that contains 3 blue pens, 2 red pens, and 3 green pens. If X is the number
of blue pens selected and Y is the number of red pens selected, find (a) the joint probability mass function, (b) P [(X, Y ) ∈ A], where A is the region {(x, y)|x + y ≤ 1}.
1 answer:
Answer:
a) f(x,y) = ; x = 0, 1 , 2; y = 0, 1 , 2; 0 ≤ x+y ≥ 2
b) =
Step-by-step explanation:
joint probability is a function that characterizes the distribution of a random variable. If X and Y be two random variables then the joint probability will be P(X = x, Y=y)
Given Data,
X = The number of blue Pens
Y = The number of red Pens
a)
possible outcomes(X, Y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), (2,0)
Please refer fig. also
total number ways of selecting any 2 pens = =
=28
f(x,y) = ; x = 0, 1 , 2; y = 0, 1 , 2; 0 ≤ x+y ≥ 2
b)
P(X,Y)∈A = P(X + Y ≤ 1)
= P(0,0) + P(1,0) + P(0,1)
= +
+
=
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<h2>Answer:</h2>
The area of the top surface of the washer is: 160.14 square mm.
<h2>Step-by-step explanation:</h2>The top of the surface is in the shape of a annulus with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)
Now, we know that the area of the annulus region is given by:
where R is the outer radius and r is the inner radius.
Here we have:
Hence, we have:
Answer:
a) 57.35%
b) 99.99%
c) 68.27%
Step-by-step explanation:
When we have a random variable X that is normally distributed with mean and standard deviation
, then
The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean and standard deviation
to the left of a.
The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean and standard deviation
to the right of b.
The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean and standard deviation
between a and b.
In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml
(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?
As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).
You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.
<em>If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula: </em>
<em>1- NORMDIST(62,63,5.4,1) in Excel </em>
<em>1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc </em>
<em>and we will get a value of 0.5735 or 57.35% </em>
(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?
Now we want the area to the left of 90
<em>NORMDIST(90,63,5.4,1) in Excel </em>
<em>NORMDIST(90;63;5.4;1) in OpenOffice Calc </em>
You will get a value of 0.9999 or 99.99%
(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?
You can use either the rule that 68.27% of the data falls between and
or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4
<em>In Excel </em>
<em>NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1) </em>
<em>In OpenOffice Calc </em>
<em>NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1) </em>
In any case we get a value of 0.6827 or 68.27%
