Measures of Dispersion

1. In summary,

The data from least to greatest is:

$2.6,3,4.9,5,5,6,6,7.9,8,8.2$

The interquartile range is 3 hours.

2. In summary,

The data from least to greatest is:

$4,4,6,7,10,11,12,14,15$

The interquartile range is 8 animal crackers.

3. 3.5

4. In summary,

The data from least to greatest is:

$0,0,0,1,2,2,4,5,7,8,10$

The interquartile range is 7 remote controllers.

1. Gabriella asked $5$ of her hundreds of coworkers how much storage space they were currently using on their computer. Here are their responses (in gigabytes):

$4,8,8,9,11$

The mean of these amounts is $\bar{x}=8$ gigabytes.

What is the standard deviation?

Round your answer to two decimal places.

$s_{x} \approx \text{_______}$ gigabytes

2. Jordyn asked $8$ people how many apps they each had on their phone. Here are their responses:

$21,28,31,46,55,60,65,70$

The mean is $\bar{x}=47$ apps.

Which of these formulas gives the standard deviation?

Choose 1 answer:

(A) $s_{x}=\sqrt{\dfrac{(21-8)^{2}+(28-8)^{2}+\cdots+(70-8)^{2}}{47}}$

(B) $s_{x}=\sqrt{\dfrac{(21-47)^{2}+(28-47)^{2}+\cdots+(70-47)^{2}}{7}}$

(C) $s_{x}=\sqrt{\dfrac{(21-8)+(28-8)+\cdots+(70-8)}{47}}$

(D) $s_{x}=\sqrt{\dfrac{(21-47)+(28-47)+\cdots+(70-47)}{7}}$

(E) $s_{x}=\dfrac{21+28+\cdots+65+70}{8}$

3. Jerry took a sample of 4 employees in his office and observed how many hours they each worked one day. Here is what he found:

$\begin{array}{lrrrr}\text { Employee } & \text { Leslie } & \text { April } & \text { Tom } & \text { Andy } \\\hline \text { Hours } & 10 & 2 & 4 & 8\end{array}$

Jerry found their mean was $\bar{x}=6$ hours. He thinks the standard deviation is

$s_{x}=\sqrt{\frac{(10-6)^{2}+(2-6)^{2}+(4-6)^{2}+(8-6)^{2}}{5}}$

What is the error in Jerry's standard deviation calculation?

Choose 1 answer:

(A) He shouldn't take the square root at all.

(B) He should only take the square root of the numerator.

(C) His denominator is incorrect.

(D) He shouldn't square each deviation in the numerator.

(E) There is no error; his calculation is correct.

4. Walker took a sample of 5 of his classmates and timed them as they completed a maze he created. Here are their times (in minutes):

$1,6,6,7,10$

The mean of these times is $\bar{x}=6$ minutes.

What is the standard deviation?

Round your answer to two decimal places.

$s_{x} \approx \text{______}$ minutes

1. The standard deviation of the amounts is $s_{x} \approx 2.55$ gigabytes.

2. The standard deviation is

$s_{x}=\sqrt{\dfrac{(21-47)^{2}+(28-47)^{2}+\cdots+(70-47)^{2}}{7}}\$

3. Jerry's error is that his denominator is incorrect.

4. The standard deviation of the times is $s_{x} \approx 3.24$ minutes.