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3 years ago

7

The dot plots below show the scores for a group of students for two rounds of a quiz: Two dot plots are shown one below the othe

r. The title for the dot plot on the top is Round 1 and the title for the bottom plot is Round 2. Below the line for each dot plot is written Score. There are markings from 1 to 5 on the line at intervals of one. There are 2 dots above the mark 2, there are 3 dots above the mark 3, there are 2 dots above the mark 4 and there is 1 dot above the mark 5. For the bottom dot plot there are there are 4 dots above the mark 4 and there are 4 dots above the mark 5. Which of the following inferences can be made using the dot plots? There is no overlap between the data. Round 2 scores are lower than round 1 scores. Round 1 scores are higher than round 2 scores. The range of round 1 is greater than the round 2 range.

2 answers:

cluponka [151]3 years ago

7 0

Answer:

The correct option is 4. The range of round 1 is greater than the round 2 range.

Step-by-step explanation:

In the given dot plots the lines represents the scores and the dots above the mark represents its frequency.

For Round 1, the frequency table is:

Score Frequency

1 0

2 2

3 3

4 2

5 1

Total score of round 1 is

$2(2)+3(3)+4(2)+5(1)=26$

From the above table it is clear that the minimum score for round 1 is 2 and the maximum score is 5.

$Range=Maximum-Minimum=5-2=3$

Therefore the range of round 1 is 3.

For Round 2, the frequency table is:

Score Frequency

1 0

2 0

3 0

4 4

5 4

Total score of round 2 is

$4(4)+5(4)=36$

From the above table it is clear that the minimum score for round 2 is 4 and the maximum score is 5.

$Range=Maximum-Minimum=5-4=1$

Therefore the range of round 2 is 1.

There is overlap between the data. Therefore option 1 is incorrect.

Since 36>26, it means the scores of round 2 are higher than round-1, therefore options 2 and 3 are incorrect.

Range of round 1 is 3 and the range of round 2 is 1, therefore the range of round 1 is greater than the round 2 range.

Hence option 4 is correct.

miskamm [114]3 years ago

6 0

Answer:

The inference that can be made using the dot plot is:

The range of round 1 is greater than the round 2 range.

Step-by-step explanation:

<u>Round 1:</u>

Score Frequency

1 0

2 2

3 3

4 2

5 1

Hence, the minimum score of Round 1 is: 2

maximum score is: 5

Hence, Range=Maximum value-Minimum score

=5-2

=3

Similarly, <u>Round-2</u>

Score Frequency

1 0

2 0

3 0

4 4

5 4

Hence, the minimum score of Round 1 is: 4

maximum score is: 5

Hence, Range=Maximum value-Minimum score

=5-4

=1

The scores of round 2 are higher than round-1.

Since round 2 have a higher frequency for higher scores as compared to round-1.

Hence, Range of round 1 is greater than the range of Round-2.

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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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andrew-mc [135]

Answer:

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Step-by-step explanation:

Given that both trjangles are similar, therefore, it follows that their corresponding side lengths would be proportional to each other. Thus:

$\frac{40}{8} = \frac{60}{2x - 2}$

$5 = \frac{60}{2x - 2}$

Cross multiply

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$10x - 10 = 60$

$10x = 60 + 10$ (addition property of equality)

$10x = 70$

Divide both sides by 10

x = 7

✔️NR = 2x - 2

Plug in the value of x

NR = 2(7) - 2 = 14 - 2

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