Respan Dataset Explorer



How many ways can a hiring committee select 3 of 12 job applicants for further consideration?

To determine the number of ways a hiring committee can select 3 out of 12 job applicants for further consideration, we use the concept of combinations. The combination formula is used when the order of selection does not matter.

The formula for combinations is:

\[
C(n, k) = \frac{n!}{k!(n - k)!}
\]

where:
- \( n \) is the total number of items (in this case, the total number of job applicants).
- \( k \) is the number of items to choose (in this case, the number of applicants to select).
- \( ! \) denotes factorial, which is the product of all positive integers up to that number.

Plugging in the values, we have:

\[
C(12, 3) = \frac{12!}{3!(12 - 3)!} = \frac{12!}{3! \cdot 9!}
\]

Since \( 12! = 12 \times 11 \times 10 \times 9! \), we can cancel the \( 9! \) in the numerator and the denominator:

\[
\frac{12 \times 11 \times 10 \times 9!}{3! \times 9!} = \frac{12 \times 11 \times 10}{3!}
\]

Now, we calculate \( 3! \):

\[
3! = 3 \times 2 \times 1 = 6
\]

So, the expression becomes:

\[
\frac{12 \times 11 \times 10}{6}
\]

Now, we perform the multiplication and division:

\[
12 \times 11 = 132
\]
\[
132 \times 10 = 1320
\]

Finally:

\[
\frac{1320}{6} = 220
\]

Thus, there are 220 different ways for the hiring committee to select 3 of the 12 job applicants for further consideration.



A pair of standard dice are rolled. Find the probability of rolling a sum of 4 with these dice.
P(D1 + D2 = 4) =
[ ? ]

To find the probability of rolling a sum of 4 with a pair of standard dice, we first determine the number of possible outcomes that result in a sum of 4, and then divide that by the total number of possible outcomes when rolling two dice.

A standard six-sided die has faces numbered from 1 to 6. When rolling two dice, the pair \((D1, D2)\) where \( D1 \) and \( D2 \) are the results on the first and second die respectively, produces a total of \(6 \times 6 = 36\) possible outcomes.

Next, we identify all pairs \((D1, D2)\) that sum to 4:

\[
\begin{aligned}
&D1 = 1, D2 = 3 \quad (1 + 3 = 4) \\
&D1 = 2, D2 = 2 \quad (2 + 2 = 4) \\
&D1 = 3, D2 = 1 \quad (3 + 1 = 4) \\
\end{aligned}
\]

There are 3 such pairs that result in a sum of 4:

1. \((1, 3)\)
2. \((2, 2)\)
3. \((3, 1)\)

Thus, there are 3 favorable outcomes.

The probability of rolling a sum of 4 with a pair of standard dice is the number of favorable outcomes divided by the total number of possible outcomes:

\[
P(D1 + D2 = 4) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{36} = \frac{1}{12}
\]

Therefore,

\[
P(D1 + D2 = 4) = \frac{1}{12}
\]



A pair of standard dice are rolled. Find the probability of rolling a sum of 11 with these dice.
P(D1 + D2 = 11) =
[?
[ ? ]

To find the probability of rolling a sum of 11 with a pair of standard dice, we first determine the number of possible outcomes that result in a sum of 11, and then divide that by the total number of possible outcomes when rolling two dice.

A standard six-sided die has faces numbered from 1 to 6. When rolling two dice, there are a total of \(6 \times 6 = 36\) possible outcomes.

Next, we identify all pairs \((D1, D2)\) that sum to 11:

\[
\begin{aligned}
&D1 = 5, D2 = 6 \quad (5 + 6 = 11) \\
&D1 = 6, D2 = 5 \quad (6 + 5 = 11) \\
\end{aligned}
\]

There are 2 such pairs that result in a sum of 11:

1. \((5, 6)\)
2. \((6, 5)\)

Thus, there are 2 favorable outcomes.

The probability of rolling a sum of 11 with a pair of standard dice is the number of favorable outcomes divided by the total number of possible outcomes:

\[
P(D1 + D2 = 11) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{36} = \frac{1}{18}
\]

Therefore:

\[
P(D1 + D2 = 11) = \frac{1}{18}
\]



Here are the last 13 ticket totals Officer Clark has handed out:
123, 65, 48, 210, 158, 67, 117, 89, 79, 156, 198, 49, 145
What is the third quartile?

To find the third quartile (Q3) of a data set, you need to follow these steps:

1. **Order the data set from smallest to largest.**
2. **Determine the position of Q3.**

The third quartile, or Q3, is the value that separates the top 25% of the data from the bottom 75%. It is also known as the 75th percentile.

Given the ticket totals:
\[ 123, 65, 48, 210, 158, 67, 117, 89, 79, 156, 198, 49, 145 \]

**Step 1: Order the data set**
\[
48, 49, 65, 67, 79, 89, 117, 123, 145, 156, 158, 198, 210
\]

**Step 2: Determine the position of Q3**

To find Q3, we use the formula:
\[ Q3 = \text{position of the } \left(\frac{3(n+1)}{4}\right) \text{th value} \]

where \( n \) is the number of data points. Here, \( n = 13 \):

\[
Q3 = \frac{3(13 + 1)}{4} = \frac{3 \times 14}{4} = \frac{42}{4} = 10.5
\]

So, Q3 is the value at the \( 10.5 \)th position in the ordered data set. This means we take the average of the values at the 10th and 11th positions.

**Step 3: Identify the 10th and 11th positions in the ordered data set**
\[
\begin{aligned}
&\text{10th value} = 156 \\
&\text{11th value} = 158 \\
\end{aligned}
\]

**Step 4: Calculate Q3**

\[
Q3 = \frac{156 + 158}{2} = \frac{314}{2} = 157
\]

Therefore, the third quartile (Q3) of the given data set is:

\[
Q3 = 157
\]

what is the median

To find the median of the given data set, we need to follow these steps:

1. **Order the data set from smallest to largest.**
2. **Find the middle value.**

Given the ticket totals:
\[ 123, 65, 48, 210, 158, 67, 117, 89, 79, 156, 198, 49, 145 \]

**Step 1: Order the data set**

\[
48, 49, 65, 67, 79, 89, 117, 123, 145, 156, 158, 198, 210
\]

**Step 2: Find the middle value**

Since the number of data points (\( n \)) is 13 (which is odd), the median is the value at the \(\left(\frac{n+1}{2}\right)\)th position.

\[
\text{Median position} = \frac{13 + 1}{2} = \frac{14}{2} = 7
\]

**Step 3: Identify the value at the 7th position in the ordered data set**

The 7th value in the ordered data set is:

\[
117
\]

Therefore, the median of the given data set is:

\[
\text{Median} = 117
\]



Find the mean, X, of the data.
16, 13, 20, 16, 12, 17, 18

To find the mean (or average) of a data set, you sum all the data values and then divide by the number of values.

Given the data set:
\[ 16, 13, 20, 16, 12, 17, 18 \]

**Step 1: Sum the data values**

\[
16 + 13 + 20 + 16 + 12 + 17 + 18 = 112
\]

**Step 2: Divide by the number of values**

There are 7 data points in the set, so we divide the sum by 7:

\[
\text{Mean} = \frac{112}{7} = 16
\]

Therefore, the mean \(X\) of the given data set is:

\[
X = 16
\]

what is the variance rounded to the nearest tenth

To find the variance of the data set, follow these steps:

1. **Calculate the mean (average) of the data set.**
2. **Subtract the mean and square the result for each data point.**
3. **Find the average of these squared differences (this is the variance).**

Given the data set:
\[ 16, 13, 20, 16, 12, 17, 18 \]

We already calculated the mean \( \mu \) to be 16.

**Step 1: Calculate the mean (already done)**
\[ \mu = 16 \]

**Step 2: Calculate each squared difference from the mean**

\[
\begin{aligned}
&(16 - 16)^2 = 0^2 = 0 \\
&(13 - 16)^2 = (-3)^2 = 9 \\
&(20 - 16)^2 = 4^2 = 16 \\
&(16 - 16)^2 = 0^2 = 0 \\
&(12 - 16)^2 = (-4)^2 = 16 \\
&(17 - 16)^2 = 1^2 = 1 \\
&(18 - 16)^2 = 2^2 = 4 \\
\end{aligned}
\]

**Step 3: Find the sum of these squared differences**

\[
0 + 9 + 16 + 0 + 16 + 1 + 4 = 46
\]

**Step 4: Divide by the number of data points to find the variance \( \sigma^2 \)**

Since we are dealing with a sample of data (and not an entire population), the sample variance \( s^2 \) is calculated using \( n - 1 \) where \( n \) is the number of data points.

\[
s^2 = \frac{\sum (X_i - \mu)^2}{n - 1}
\]

\(\sum (X_i - \mu)^2 = 46\)

\(n = 7\)

\[
s^2 = \frac{46}{7-1} = \frac{46}{6} \approx 7.67
\]

Therefore, the variance of the data set rounded to the nearest tenth is:

\[
s^2 \approx 7.7
\]

what is the standard deviation

To find the standard deviation of the data set, you first need to calculate the variance (which we already did), and then take the square root of that variance. 

Given the data set:
\[ 16, 13, 20, 16, 12, 17, 18 \]

We had already calculated the variance \( s^2 \approx 7.67\).

The standard deviation is the square root of the variance. 

\[
s = \sqrt{s^2} = \sqrt{7.67}
\]

Now, calculate the square root of 7.67:

\[
s \approx 2.77
\]

Therefore, the standard deviation of the data set is approximately:

\[
s \approx 2.8
\]

(Note: While the exact variance computed from \(\frac{46}{6}\) was used, the standard deviation remains 2.77 which rounds to 2.8 when rounded to the nearest tenth.)



Find the mean, X, of the data.
86, 63, 96, 80, 90

To find the mean (or average) of the given data set, sum all the data values and then divide by the number of values.

Given the data set:
\[ 86, 63, 96, 80, 90 \]

**Step 1: Sum the data values**

\[
86 + 63 + 96 + 80 + 90 = 415
\]

**Step 2: Divide by the number of values**

There are 5 data points in the set, so we divide the sum by 5:

\[
\text{Mean} = \frac{415}{5} = 83
\]

Therefore, the mean \(X\) of the given data set is:

\[
X = 83
\]

what is the variance and standard deviation

To find the variance and standard deviation of the given data set, follow these steps:

**Given the data set:**
\[ 86, 63, 96, 80, 90 \]

**Step 1: Calculate the mean**

The mean \( \mu \) has already been calculated as 83.

**Step 2: Calculate each squared difference from the mean**

\[
\begin{aligned}
&(86 - 83)^2 = 3^2 = 9 \\
&(63 - 83)^2 = (-20)^2 = 400 \\
&(96 - 83)^2 = 13^2 = 169 \\
&(80 - 83)^2 = (-3)^2 = 9 \\
&(90 - 83)^2 = 7^2 = 49 \\
\end{aligned}
\]

**Step 3: Find the sum of these squared differences**

\[
9 + 400 + 169 + 9 + 49 = 636
\]

**Step 4: Divide by the number of data points to find the variance**

Since we are calculating for a sample (and not an entire population), we use \( n - 1 \) where \( n \) is the number of data points.

\[
s^2 = \frac{\sum (X_i - \mu)^2}{n - 1}
\]

\[
s^2 = \frac{636}{5 - 1} = \frac{636}{4} = 159
\]

**Step 5: Calculate the standard deviation**

The standard deviation \( s \) is the square root of the variance:

\[
s = \sqrt{159} \approx 12.61
\]

So, the variance and standard deviation of the data set are:

\[
\text{Variance } (s^2) = 159
\]

\[
\text{Standard Deviation } (s) \approx 12.61
\]