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# Statistics Worksheets

Online Statistics Worksheets
Statistics Worksheets
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1.
The table shows the average time (in hours) that a driver is delayed due to road congestion in a year at different places.
 Place Number of hours A 53 B 46 C 44 D 42 E 36
Find Q1, Q2, Q3 for this data.
 a. Q1 = 44, Q2 = 42 and Q3 = 46 b. Q1 = 39, Q2 = 44 and Q3 = 49.5 c. Q1 = 44, Q2 = 39 and Q3 = 49.5 d. Q1 = 36, Q2 = 44 and Q3 = 53

#### Solution:

Arrange the number of hours in increasing order.
36, 42, 44, 46, 53

Median = 44.
[Median is the middle value of the data set in the order.]

So, Q2 = 44
[Q2 is the median.]

The median of the data values less than 44 (median) is Q1.

The number of hours that are less than 44 (median) are 36, 42.

Q1 = 36+42 / 2 = 39

The median of the data values greater than 44 (median) is Q3.

The number of hours that are greater than 44 (median) are 46, 53.

Q3 = 46+53 / 2 = 49.5

So, Q1 = 39, Q2 = 44 and Q3 = 49.5.

2.
For a group of 200 candidates, the mean and standard deviation of scores were found to be 40 and 15. Later on, it was discovered that the scores 43 and 35 were misread as 34 and 53. Find the corrected mean and standard deviation.
 a. 39.955 and 14.97 b. 31 and 6 c. 39.955 and 224.14 d. 39.955 and 42.66

#### Solution:

Let the sum of the scores be x.

40 = x200
x = 40 × 200 = 8000
[Mean = sum of the termsnumber of terms.]

Corrected sum, x = [8000 - (34 + 53) + 43 + 35] = 7991.

Original mean, X = 40 and standard deviation, σ = 15 then variance, σ2 = 152 = 225.

x2 = n2 + x2) = 200(225 + 1600) = 365000
[Variance = (σ)2 = 1n x2 - x2.]

Corrected x2 = 365000 - (34)2 - (53)2 + (43)2 + (35)2 = 364109

Corrected variance, σ2 = 364109 / 200 - (39.955)2 = 1820.54 - 1596.40 = 224.14

Corrected standard deviation(σ) = 224.14 = 14.97.

So, the corrected mean and standard deviation of the scores are 39.955 and 14.97.

3.
When a distribution is bell-shaped, approximately what percentage of data values will fall within 1 standard deviation of the mean?
I. 95%
II. 99.7%
III. 68%
 a. only III b. only I c. only II d. all are correct

#### Solution:

When a distribution is bell - shaped, approximately 95% of the data values will fall within 2 standard deviations of the mean.

When a distribution is bell - shaped, approximately 99.7% of the data values will fall within 3 standard deviations of the mean.

When a distribution is bell - shaped, approximately 68% of the data values will fall within 1 standard deviation of the mean.

4.
Which among the following statement(s) is/are correct?
I. An outlier is an extremely high or an extremely low data value when compared with the rest of the data values.
II. The mean and standard deviation of a variable are not affected by an outlier.
III. The interquartile range is defined as the difference between Q1 and Q3 and is the range of the middle 25% of the data.
IV. Q2 is same as the mean.
 a. I only b. II and IV only c. II and III only d. III only

#### Solution:

An outlier is an extremely high or an extremely low data value when compared with the rest of the data values.

An outlier can strongly affect the mean and standard deviation of a variable.

The interquartile range(IQR) is defined as the difference between Q1 and Q3 and is the range of the middle 50% of the data.

Q2 is same as the 50th percentile or median.

So, only statement I is correct.

5.
A partial list of U.S Presidents and their respective ages when they took the charge of the White House are given. Find the five-number summary. Was Bill Clinton in the youngest 25% for the given list?
 President Age Washington 57 Jackson 61 W.H.Harrison 68 Lincoln 52 A.Johnson 56 B.Harison 55 J.Roosevelt 42 F.D.Roosevelt 51 Kennedy 43 Clinton 46

 a. 46, 53.5, 57 and yes b. 42, 46, 53.5, 57, 68 and yes c. 42, 44.5, 53.5, 59, 68 and no d. 42, 44, 53, 59, 68 and no

#### Solution:

Arrange the data in order: 42, 43, 46, 51, 52, 55, 56, 57, 61, 68.

The values
1. lowest value of the data set,
2. Q1,
3. median,
4. Q3,
5. highest value of the data set
are called the five-number summary of the data set.

Median quartile, Q2 = 52+552 = 53.5

Lower quartile, Q1 is the median of the data values that fall below Q2.

Q1 = 46

Upper quartile, Q3 is the median of the data values that fall above Q2.

Q3 = 57

The five-number summary is 42, 46, 53.5, 57, 68.

Age of Bill Clinton when he took office is 46 = Q1, hence Bill Clinton falls in the youngest 25%.

6.
Select the true statement(s).
I. The measure of central tendency used in exploratory data analysis is mean.
II. The measure of central tendency used in $z$ score is median.
III. The mean and standard deviation is better than five-number summary for describing a skewed distribution.
 a. I only b. II only c. III only d. all statements are wrong

#### Solution:

The measure of central tendency used in exploratory data analysis is median.

The measure of central tendency used in z score is mean.

The five-number summary is better than the mean and standard deviation for describing a skewed distribution.

So, all statements are wrong.

7.
The reaction time to a stimulus for a certain drug has a mean of 2.5 seconds and a standard deviation of 0.3 seconds. Find the $z$ score for a reaction time of 3.1 seconds.
 a. 2 b. 18.67 c. 0.2 d. 7.33

#### Solution:

z = value - meanstandard deviation= X-Xs

= 3.1-2.5 / 0.3 = 0.6 / 0.3 = 2

So, the z score for a reaction time of 3.1 seconds is 2.

8.
A talent test has a mean of 220 and a standard deviation of 10. Which among the following statement(s) is/are true?
I. If the $z$ score for an exam score is negative, then the score is less than 220.
II. If the $z$ score for an exam score is zero, then the score is equal to 220.
III. If the $z$ score is positive, then the score is more than 220.
 a. II only b. I and III only c. III only d. I, II and III

#### Solution:

z score for the test, z = X-Xs, where X is the score, X the mean and s the standard deviation of the scores in the test.

z = X-22010

If the z score of an exam is negative, then
X-22010 < 0

X < 220

Tha statement I is true.

If the z score of an exam is zero, then
X-22010 = 0

X = 220

Tha statement II is true.

If the z score of an exam is positive, then
X-22010 > 0

X > 220

Tha statement III is true.

9.
Jessica scored 70 in an Algebra test that had a mean of 60 and a standard deviation of 8, and she scored 56 in French with a mean of 50 and a standard deviation of 5. Compare her relative positions in the two tests.
 a. position in algebra is higher b. position in French is higher c. position in both algebra and French are same d. cannot be determined

#### Solution:

The z score for Algebra test, z = X-Xs = 70-60 / 8 = 10 / 8 = 1.25

The z score for French test, z = X-xs = 56-50 / 5 = 6 / 5 = 1.2

Since the z score for Algebra is larger, her relative position in the Algebra is higher than the relative position in French.