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## Question ac7d4

Jason K.
Featured 4 months ago

68.53% of bearings can be used on the machine.

#### Explanation:

The normal distribution described by this problem is $N \left(0.400 , {0.001}^{2}\right)$.

The range of acceptable bearing sizes in this problem is given by $0.399 \pm 0.0015$. That equates to the interval $\left[0.3975 . 0.4005\right]$.

To determine the percentage, it will be helpful for us to convert these two endpoints from this normal distribution into z-scores on the standard normal distribution ${N}_{s} \left(0 , {1}^{2}\right)$. We can then use z-score tables (or a z-score calculator) to determine what the area under the standard normal distribution curve is between those two z-scores.

Recall the z-score formula:

$z = \frac{x - \mu}{\sigma}$

Thus:

${z}_{0.3975} = {z}_{1} = \frac{0.3975 - 0.400}{0.001} = - 2.5$

${z}_{0.4005} = {z}_{2} = \frac{0.4005 - 0.400}{0.001} = 0.5$

Next, we turn to a z-score table to find the cumulative from $- \infty$ area measures for each of the z-scores. Once we have those, we simply subtract ${z}_{1}$ from ${z}_{2}$ to find the area between those two z-scores:

$P \left(Z < {z}_{1}\right) = P \left(Z < - 2.5\right) = 0.0062$

$P \left(Z < {z}_{2}\right) = P \left(Z < 0.5\right) = 0.6915$

{:(P(z_1 < Z < z_2),=, P(-2.5 < Z < 0.5)), (,=, P(Z < 0.5) - P(Z < -2.5)), (,=,0.6915-0.0062), (,=,0.6853):}

68.53% of bearings can be used on the machine.

## What is the interquartile range of the set of data: 67, 58, 79, 85, 80, 72, 75, 76, 59, 55, 62, 67, 80?

Jason K.
Featured 4 months ago

$I Q R = 19$
(Or 17, see note at end of explanation)

#### Explanation:

The interquartile range (IQR) is the difference between the 3rd Quartile value (Q3) and the 1st Quartile value (Q1) of a set of values.

To find this, we need to first sort the data in ascending order:

55, 58, 59, 62, 67, 67, 72, 75, 76, 79, 80, 80, 85

Now we determine the median of the list. The median is generally known as the number is the "center" of the ascending ordered list of values. For lists with an odd number of entries, this is easy to do as there is a single value for which an equal number of entries are less than or equal and greater than or equal. In our sorted list, we can see that the value 72 has exactly 6 values less than it and 6 values greater than it:

$\textcolor{b l u e}{55 , 58 , 59 , 62 , 67 , 67 ,} \textcolor{red}{72 ,} \textcolor{g r e e n}{75 , 76 , 79 , 80 , 80 , 85}$

Once we have the median (also sometimes referred to as the 2nd Quartile [Q2]), we can determine the Q1 and Q3 by finding the medians of the lists of values below and above the median, respectively.

For Q1, our list (colored in blue above) is 55, 58, 59, 62, 67, and 67. There is an even number of entries in this list, and therefore a common convention to use for finding the median in an even list is to take the two "center most" entries in the list and find their mean [arithmetic average]. Thus:

$Q 1 = \frac{59 + 62}{2} = \frac{121}{2} = 60.5$

For Q2, our list (colored in green above) is 75, 76, 79, 80, 80, and 85. Again, we will find the mean of the two center most entries:

$Q 3 = \frac{79 + 80}{2} = 79.5$

Finally, the IQR is found by subtracting $Q 3 - Q 1$:

$I Q R = Q 3 - Q 1 = 79.5 - 60.5 = 19$

Special note:

Like many things in statistics, there are often many accepted conventions for how to calculate something. In this case, it is common for some mathematicians, when calculating Q1 and Q3 for an even number of entries (such as we did above), to actually include the median as a value in the grouping in order to avoid taking the mean of the sublists. Thus, in that case, the Q1 list would actually be 55, 58, 59, 62, 67, 67, and 72, leading to a Q1 of 62 (rather than 60.5). The Q3 would likewise be calculated to be 79 instead of 79.5, with a final IQR of 17.

## Question 0700a

Parzival S.
Featured 4 months ago

1287

#### Explanation:

Combinations use this general formula:

C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

C_(13,5)=(13!)/((5!)(13-5)!)=(13!)/((5!)(8!))

To solve, you can use a table to find the different values (or a calculator), but factorials get quite big quite fast, and so I find it's often easier to simply work out the cancellations yourself:

(13!)/((5!)(8!))=(13xxcancel12xx11xxcancel10xx9xxcancel(8!))/(cancel(5xx4xx3xx2)xxcancel(8!))=13xx11xx9=1287

This shows the number of 5-player groups we can make.

~~~~~

If we care about the order of which player goes first or about what position each player will take, then we'd want to use the permutation formula:

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

## Question 5ac9e

EZ as pi
Featured 4 months ago

Without replacement: $P \left(Y Y\right) = \frac{1}{11}$

With replacement: $P \left(Y Y\right) = \frac{1}{9}$

#### Explanation:

"Probability" = ("number of desirable outcomes")/("total number of possible outcomes")

To have BOTH balls yellow means we want to find:

P(Y Y)" larr the first ball is yellow AND the second ball is yellow.

For the first ball: P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")

If the ball is NOT replaced, there is one less yellow ball and one less ball.
For the second ball: P(Y) = 3/11 " "(larr3 " yellow balls")/(larr 11 " balls altogether")

$P \left(Y Y\right) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{11} \text{ } \leftarrow$ (AND implies multiply)

If the first ball IS replaced, the probability stays the same for the second ball being yellow.

For the second ball: P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")

$P \left(Y Y\right) = \frac{4}{12} \times \frac{4}{12} = \frac{16}{144} = \frac{1}{9} \text{ } \leftarrow$ (AND implies multiply)

## Joint probability- mean and variance? Pl refer image attached

Geoff K.
Featured 3 months ago

The answer is $\text{(d) } \frac{3}{4} {x}_{2} \mathmr{and} \frac{3}{80} {x}_{2}^{2}$.

#### Explanation:

Part 1: Conditional Mean
By definition, $\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$,

where

$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$,

where

$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{A} f \left({x}_{1} , {x}_{2}\right) {\mathrm{dx}}_{1}$.

In this case, since ${x}_{1}$ is bounded by $0 < {x}_{1} < {x}_{2}$, our integration interval is $A = \left(0 , {x}_{2}\right) .$ Thus:

$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{0}^{{x}_{2}} 21 \text{ "x_1^2" "x_2^3" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ "x_2^3int_0^(x_2)x_1^2" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ } {x}_{2}^{3} {\left[\frac{1}{3} {x}_{1}^{3}\right]}_{0}^{{x}_{2}}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 7 \text{ } {x}_{2}^{6}$

Thus,

$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$

$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \left(21 \text{ "x_1^2" "x_2^3)/(7" } {x}_{2}^{6}\right)$

$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{3 \text{ } {x}_{1}^{2}}{{x}_{2}^{3}}$.

Finally,

$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = {\int}_{A} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$

Since we are once again integrating with respect to ${x}_{1}$, the integration interval is the same as before:

$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{0}^{{x}_{2}} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$

color(white)("E"(X_1|X_2))=3/x_2^3 int_0^(x_2)x_1^3" "dx_1

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{x} _ {2}^{3} {\left[{x}_{1}^{4} / 4\right]}_{0}^{{x}_{2}}$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4 {x}_{2}^{3}} \left[{x}_{2}^{4}\right]$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4} {x}_{2}$

Part 2: Conditional Variance

The conditional variance is

"Var"(X_1|X_2)="E"(X_1^2|X_2)-["E"(X_1|X_2)]^2

I'll leave the calculation of $\text{E} \left({X}_{1}^{2} | {X}_{2}\right)$ as an exercise. (Hint: just replace ${x}_{1}$ with ${x}_{1}^{2}$ in the formula for $\text{E} \left({X}_{1} | {X}_{2}\right)$.)

The result is:

$\text{Var} \left({X}_{1} | {X}_{2}\right) = \frac{3 {x}_{2}^{2}}{5} - {\left[\frac{3 {x}_{2}}{4}\right]}^{2}$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3 {x}_{2}^{2}}{5} - \frac{9 {x}_{2}^{2}}{16}$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{1}{5} - \frac{3}{16}\right]$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{16 - 15}{80}\right]$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{80} {x}_{2}^{2}$.

## A pair of fair six-sided dice is thrown eight times. Find the probability that a score greater than 7 is scored no more than five times?

Parzival S.
Featured 2 months ago

$\cong 0.9391$

#### Explanation:

Before we get into the question itself, let's talk about the method for solving it.

Let's say, for instance, that I want to account for all the possible results from flipping a fair coin three times. I can get HHH, TTT, TTH, and HHT.

The probability of H is $\frac{1}{2}$ and the probability for T is also $\frac{1}{2}$.

For HHH and for TTT, that is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each.

For TTH and HHT, it's also $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each, but since there are 3 ways I can get each result, it ends up being $3 \times \frac{1}{8} = \frac{3}{8}$ each.

When I sum up these results, I get $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$ - which means I now have all the possible results of the coin flip accounted for.

Notice that if I set $H$ to be $p$ and therefore have $T$ be ~p, and also notice that we have a line from the Pascal's Triangle $\left(1 , 3 , 3 , 1\right)$, we've set up a form of:

sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))

and so in this example, we get:

$= {C}_{3 , 0} {\left(\frac{1}{2}\right)}^{0} {\left(\frac{1}{2}\right)}^{3} + {C}_{3 , 1} {\left(\frac{1}{2}\right)}^{1} {\left(\frac{1}{2}\right)}^{2} + {C}_{3 , 2} {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2}\right)}^{1} + {C}_{3 , 3} {\left(\frac{1}{2}\right)}^{3} {\left(\frac{1}{2}\right)}^{0}$

$= 1 \left(1\right) \left(\frac{1}{8}\right) + 3 \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) + 3 \left(\frac{1}{4}\right) \left(\frac{1}{2}\right) + 1 \left(\frac{1}{8}\right) \left(1\right)$

$= \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$

Now we can do the problem.

We're given the number of rolls as 8, so $n = 8$.

$p$ is the sum greater than 7. To find the probability of getting a sum greater than 7, let's look at the possible rolls:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

Out of 36 possibilities, 15 rolls give a sum greater than 36, giving a probability of $\frac{15}{36} = \frac{5}{12}$.

With p=5/12, ~p=7/12#

We can write out the entire sum of possibilities - from getting all 8 rolls being a sum greater than 7 all the way to getting all 8 rolls being a sum of 7 or less:

$= {C}_{8 , 0} {\left(\frac{5}{12}\right)}^{8} {\left(\frac{7}{12}\right)}^{0} + {C}_{8 , 1} {\left(\frac{5}{12}\right)}^{7} {\left(\frac{7}{12}\right)}^{1} + {C}_{8 , 2} {\left(\frac{5}{12}\right)}^{6} {\left(\frac{7}{12}\right)}^{2} + {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8} = 1$

but we're interested in summing up only those terms that have our greater than 7 sum happening 5 times or less:

$= {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8}$

$\cong 0.9391$

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