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Marta_Voda [28]
3 years ago
11

In a random sample, 10 employees at a local plant were asked to compute the distance they travel to work to the nearest tenth of

a mile. The data is listed below.
1.1, 5.2, 3.6, 5.0, 4.8, 1.8, 2.2, 5.2, 1.5, 0.8
1. Compute the range, sample standard deviation and sample variance of the data.
2. Calculate the mean, median and the mode of the data.
3. Based on these numbers, what is the shape of the distribution?
Mathematics
1 answer:
Marta_Voda [28]3 years ago
5 0

Answer:

1) Range = Max-Min= 5.2-0.8=4.4

The sample variance can be calculated with this formula:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

And we got:

s^2 = 3.324

And the sample deviation would be:

s= \sqrt{3.324}= 1.823

2) \bar X =\frac{\sum_{i=1}^n X_i}{n}

And we got:

\bar X= 3.12

The median since we have 10 values would be the average between the 5th and 6th observations from the dataset ordered and we got:

Median= \frac{2.2+3.6}{2}= 2.9

And finally the mode would be the most repated value and we got:

Mode= 5.2

3) Median

So then we can conclude that probably this distirbution is left skewed

Step-by-step explanation:

We have the following dataset given:

1.1, 5.2, 3.6, 5.0, 4.8, 1.8, 2.2, 5.2, 1.5, 0.8

Part 1

We can order the dataset on increasing way and we got:

0.8 1.1 1.5 1.8 2.2 3.6 4.8 5.0 5.2 5.2

The range can be calculated like this:

Range = Max-Min= 5.2-0.8=4.4

The sample variance can be calculated with this formula:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

And we got:

s^2 = 3.324

And the sample deviation would be:

s= \sqrt{3.324}= 1.823

Part 2

The mean can be calculated with this formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

And we got:

\bar X= 3.12

The median since we have 10 values would be the average between the 5th and 6th observations from the dataset ordered and we got:

Median= \frac{2.2+3.6}{2}= 2.9

And finally the mode would be the most repated value and we got:

Mode= 5.2

Part 3

For this case we know that:

Median

So then we can conclude that probably this distirbution is left skewed

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