Math / Data & Statistics

QuestionUsing the following data answer questions 6 \& 7 : \begin{tabular}{|c|l|} \hline Value & Frequency \\ \hline 1 & $X$ \\ \hline 2 & $X X$ \\ \hline 3 & $X X X$ \\ \hline 4 & $X X X X$ \\ \hline 5 & $X X X X X$ \\ \hline 6 & $X$ \\ \hline \end{tabular} histogram A
HISTOGRAM B
HISTOGRAM C histogRam D
6. (1 point) Which histogram represents the data?
7. (1 point) What is the skewness of the data? a. Left skew b. Right skew c. Bimodal d. Symmetric
8. (1 point) The standard deviation of $-5,-5,-5,-5,-5$ is $\qquad$ a. -5 b. 5 c. 0 d. -25
9. (1 point) The mean of 11 numbers is 7 . One of the numbers, 13 , is deleted. What is the mean of the remaining 10 numbers? a. 7.7 b. 6.4 c. 6.0

Studdy Solution

STEP 1

What is this asking? We're looking at some data presented as a frequency table, figuring out which histogram represents it, describing its skewness, and then calculating the standard deviation of a set of identical numbers and the new mean of a set of numbers after removing one value. Watch out! Histograms can be tricky!
Make sure you're matching the frequencies to the correct values.
Also, don't mix up the different types of skewness.
Remember, standard deviation measures spread, not the actual values themselves.

STEP 2

1. Match the Histogram
2. Determine Skewness
3. Calculate Standard Deviation
4. Calculate New Mean

STEP 3

Let's decode this table!
The "Value" column tells us what numbers we have, and the "Frequency" column tells us how many times each number appears.
For example, the value $1$ appears $X$ times, the value $2$ appears $XX$ times (meaning two times), and so on.

STEP 4

So, we have one 1, two 2s, three 3s, four 4s, five 5s, and one 6.
We need to find the histogram that shows these frequencies.
Look for the histogram where the bar height matches the number of times each value appears.
The correct histogram will have a bar of height 1 above the value 1, a bar of height 2 above the value 2, and so on.

STEP 5

Imagine the histogram as a mountain range.
If the mountain slopes down gradually to the right, it's a right skew.
If it slopes down gradually to the left, it's a left skew.
If it's perfectly symmetrical, it's symmetric.
If it has two distinct peaks, it's bimodal.

STEP 6

Our data has a peak at the value $5$ and then drops off sharply.
The frequencies increase up to $5$ and then decrease.
This creates a tail on the right side, so it's a right skew.

STEP 7

Standard deviation measures how spread out the data is.
If all the numbers are the same, there's no spread at all!

STEP 8

Since all the numbers are $-5$ , the standard deviation is $\mathbf{0}$ .
There's no variation in the data.

STEP 9

If the mean of $11$ numbers is $7$ , their sum is $11 \cdot 7 = \mathbf{77}$ .
This is because the mean is the sum divided by the number of values.

STEP 10

We remove $13$ , so the new sum is $77 - 13 = \mathbf{64}$ .

STEP 11

Now we have $10$ numbers.
The new mean is $64 \div 10 = \mathbf{6.4}$ .

STEP 12

6. The histogram that represents the data will have bars with heights corresponding to the frequencies: 1, 2, 3, 4, 5, and 1.
7. The skewness of the data is right skew.
8. The standard deviation is $\mathbf{0}$ .
9. The new mean is $\mathbf{6.4}$ .

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Studdy Solution

STEP 1

STEP 2

1. Match the Histogram
2. Determine Skewness
3. Calculate Standard Deviation
4. Calculate New Mean

STEP 3

Let's decode this table!
The "Value" column tells us what numbers we have, and the "Frequency" column tells us how many times each number appears.
For example, the value $1$ appears $X$ times, the value $2$ appears $XX$ times (meaning two times), and so on.

STEP 4

STEP 5

Imagine the histogram as a mountain range.
If the mountain slopes down gradually to the right, it's a right skew.
If it slopes down gradually to the left, it's a left skew.
If it's perfectly symmetrical, it's symmetric.
If it has two distinct peaks, it's bimodal.

STEP 6

Our data has a peak at the value $5$ and then drops off sharply.
The frequencies increase up to $5$ and then decrease.
This creates a tail on the right side, so it's a right skew.

STEP 7

Standard deviation measures how spread out the data is.
If all the numbers are the same, there's no spread at all!

STEP 8

Since all the numbers are $-5$ , the standard deviation is $\mathbf{0}$ .
There's no variation in the data.

STEP 9

If the mean of $11$ numbers is $7$ , their sum is $11 \cdot 7 = \mathbf{77}$ .
This is because the mean is the sum divided by the number of values.

STEP 10

We remove $13$ , so the new sum is $77 - 13 = \mathbf{64}$ .

STEP 11

Now we have $10$ numbers.
The new mean is $64 \div 10 = \mathbf{6.4}$ .

STEP 12

6. The histogram that represents the data will have bars with heights corresponding to the frequencies: 1, 2, 3, 4, 5, and 1.
7. The skewness of the data is right skew.
8. The standard deviation is $\mathbf{0}$ .
9. The new mean is $\mathbf{6.4}$ .

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Studdy Solution

STEP 1

STEP 2

1. Match the Histogram2. Determine Skewness3. Calculate Standard Deviation4. Calculate New Mean

STEP 3

Let's **decode** this table!The "Value" column tells us what numbers we have, and the "Frequency" column tells us how many times each number appears.For example, the value 111 appears XXX times, the value 222 appears XXXXXX times (meaning two times), and so on.

STEP 4

STEP 5

Imagine the histogram as a mountain range.If the mountain slopes down gradually to the right, it's a **right skew**.If it slopes down gradually to the left, it's a **left skew**.If it's perfectly symmetrical, it's **symmetric**.If it has two distinct peaks, it's **bimodal**.

STEP 6

Our data has a peak at the value 555 and then drops off sharply.The frequencies increase up to 555 and then decrease.This creates a tail on the right side, so it's a **right skew**.

STEP 7

Standard deviation measures how spread out the data is.If all the numbers are the same, there's *no spread* at all!

STEP 8

Since all the numbers are −5-5−5, the standard deviation is 0\mathbf{0}0.There's no variation in the data.

STEP 9

If the mean of 111111 numbers is 777, their sum is 11⋅7=7711 \cdot 7 = \mathbf{77}11⋅7=77.This is because the mean is the sum divided by the number of values.

STEP 10

We **remove** 131313, so the new sum is 77−13=6477 - 13 = \mathbf{64}77−13=64.

STEP 11

Now we have 101010 numbers.The new mean is 64÷10=6.464 \div 10 = \mathbf{6.4}64÷10=6.4.

STEP 12

6. The histogram that represents the data will have bars with heights corresponding to the frequencies: 1, 2, 3, 4, 5, and 1.7. The skewness of the data is **right skew**.8. The standard deviation is 0\mathbf{0}0.9. The new mean is 6.4\mathbf{6.4}6.4.

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1. Match the Histogram
2. Determine Skewness
3. Calculate Standard Deviation
4. Calculate New Mean

Let's decode this table!
The "Value" column tells us what numbers we have, and the "Frequency" column tells us how many times each number appears.
For example, the value $1$ appears $X$ times, the value $2$ appears $XX$ times (meaning two times), and so on.

Imagine the histogram as a mountain range.
If the mountain slopes down gradually to the right, it's a right skew.
If it slopes down gradually to the left, it's a left skew.
If it's perfectly symmetrical, it's symmetric.
If it has two distinct peaks, it's bimodal.

Our data has a peak at the value $5$ and then drops off sharply.
The frequencies increase up to $5$ and then decrease.
This creates a tail on the right side, so it's a right skew.

Standard deviation measures how spread out the data is.
If all the numbers are the same, there's no spread at all!

Since all the numbers are $-5$ , the standard deviation is $\mathbf{0}$ .
There's no variation in the data.

If the mean of $11$ numbers is $7$ , their sum is $11 \cdot 7 = \mathbf{77}$ .
This is because the mean is the sum divided by the number of values.

We remove $13$ , so the new sum is $77 - 13 = \mathbf{64}$ .

Now we have $10$ numbers.
The new mean is $64 \div 10 = \mathbf{6.4}$ .

6. The histogram that represents the data will have bars with heights corresponding to the frequencies: 1, 2, 3, 4, 5, and 1.
7. The skewness of the data is right skew.
8. The standard deviation is $\mathbf{0}$ .
9. The new mean is $\mathbf{6.4}$ .