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Poisson Distribution Examples

Poisson Distribution Examples
• Page 1
1.
The number of years served by 9 Prime Ministers of a country are 3, 5, 4, 6, 2, 7, 8, 1 and 10. Find the median and the midrange of the above data. Are they equal?
 a. 5.5, 5.5, yes b. 5, 5.5, no c. 5, 5, yes d. 5.11, 5, no

Solution:

To find median, arrange the data values in order.

1, 2, 3, 4, 5, 6, 7, 8, 10

Median = 5
[Middle value of the data set is the median.]

Midrange = lowest value + highest value2 = 1+102 = 11 / 2 = 5.5

So, the median is not equal to the midrange of the length of serivce offered by different prime ministers.

Correct answer : (2)
2.
A sales person drives 200 miles round trip at 25 mph going to California and 40 mph returning home. Find the average speed.
 a. 30.8 mph b. 3.08 mph c. 15.4 mph d. 6.1 mph

Solution:

The average speed is the harmonic mean of the two speeds.

H.M = 2125+140 = 2(200)8+5 = 400 / 13 = 30.8 mph.
[H.M = nΣ(1X).]

So, the average speed of the sales person is 30.8 mph.

Correct answer : (1)
3.
Find the mean of 5, 10, 15, 20, 25 and 30. Increase each value by 15 and find the mean of the new data. Are the two means the same?
 a. 17.5, 17.5, yes b. 17.5, 32.5, no c. 17.5, 35, no d. 17.5, 32.5, yes

Solution:

X1 = ∑X / n = 5+10+15+20+25+306 = 105 / 6 = 17.5

Add 15 to each data value in the given data set. The new data set will be 20, 25, 30, 35, 40, 45.

New mean (X2) = ∑X / n = 20+25+30+35+40+456 = 32.5

= 17.5 + 15

= X1 + the number added to each data value in the given data set.

The measures of the two means are not the same.

Correct answer : (2)
4.
The time (in minutes) taken by 10 students to complete a statistics test was 15, 10, 12, 7, 9, 13, 15, 6, 14, 8. Find the midrange and comment on the skewness of the distribution.
 a. 11, symmetric b. 11, left skewed c. 10.5, right skewed d. 10.5, left skewed

Solution:

15, 10, 12, 7, 9, 13, 15, 6, 14, 8
Mean, X = ∑x / n = 15+10+12+7+9+13+15+6+14+810 = 10.9

So, mean time of 10 students to complete the statistics test is 10.9 minutes.

Arrange the above data values in order: 6, 7, 8, 9, 10, 12, 13, 14, 15, 15.

Median = 10+122 = 11
[Median is the average of middle values in the data set.]

So, median is 11 minutes.

Since 2 students have taken 15 minutes to complete the test, the mode for the data set is 15. This data set is said to be uni-modal.

= 6+152 = 10.5
Midrange = lowestvalue+highestvalue2

Arrange Mean, Median, Mode and Midrange in ascending order.

10.5 < 10.9 < 11 < 15 Midrange < Mean < Median < Mode

The majority of data values fall to the right of the mean and cluster at the upper end of the distribution, with the tail to the left. So, this distribution is negatively skewed or left skewed.

Correct answer : (4)
5.
The number of ATM transactions of a particular bank recorded at 12 locations in a city on a day are as follows.
24, 42, 210, 69, 38, 72, 59, 44, 96, 21, 318, 66
Find the median and mean of the number of transactions. Which is greater, median or mean?
 a. 62.5, 88.25, mean b. 66, 88.25, mean c. 88.25, 62.5, median d. 169.5, 88.25, median

Solution:

Arrange the given data in order: 21, 24, 38, 42, 44, 59, 66, 69, 72, 96, 210, 318

Median = 59+662 = 62.5
[Median is the average of middle values of the data set.]

Mean, X = ∑X / n = 24 + 42 + 210 + 69 + 38 + 72 + 59 + 44 + 96 + 21 + 318 + 6612

= 1059 / 12 = 88.25

Mean is greater than median.
[Compare the Mean and median of the data.]

Correct answer : (1)
6.
The data represents the percentage of children suffering from vascular diseases in 8 selected areas of 2 different countries. Find the mean, median, mode and midrange for country 1 and country 2.

 a. Country 1: 8.263%, 7.92%, 2.9%, 8.225% ; Country 2: 6.116%, 3.88%, 2.9%, 6.555% b. Country 1: 6.116%, 7.92%, no mode, 8.225% ; Country 2: 8.263%, 3.88%, 2.9%, 6.555% c. Country 1: 8.263%, 7.92%, no mode, 8.225% ; Country 2: 6.116%, 3.88%, 2.9%, 6.555% d. Country 1: 8.263%, 7.92%, no mode, 8.225% ; Country 2: 6.116%, 3.88%, no mode, 6.555%

Solution:

Mean percentage of children suffering from vascular diseases in country 1 is X = ∑X / n

= 5+6.83+2.9+14.56+9.01+1.89+12.91+138

= 66.18 = 8.263%

So, the mean percentage of children suffering from vascular diseases in country 1 = 8.263%.

Arrange the data values of country 1 in order: 1.89%, 2.9%, 5%, 6.83%, 9.01%, 12.91%, 13%, 14.56%.

The median percentage is MD = 6.83+9.012 = 7.92%.

There is no mode since each value occurs only once.

Midrange, MR = 1.89+14.562 = 8.225%.
[Use Mid range = lowest value+highest value2]

Mean percentage of children suffering from vascular diseases in country 2 is X = ∑X / n

= 2.9+1.02+12.91+13.02+4.86+2.9+0.09+11.238

= 48.93 / 8 = 6.116%

So, the mean percentage of children suffering from vascular diseases in country 2 = 6.116%.

Arrange the data values of country 2 in order: 0.09%, 1.02%, 2.9%, 2.9%, 4.86%, 11.23%, 12.91%, 13.02%.

The median percentage is MD = 2.9+4.862 = 3.88%.

The mode is 2.9 since 2.9 occurs 2 times.

Midrange, MR = 0.09+13.022 = 6.555%.
[Use Mid range = lowest value+highest value2]

Correct answer : (3)
7.
A recent survey of different companies reported the following percentages of people working in night shifts. Find the weighted mean of percentages.
 Company % of people working in night shifts ($w$) Number of people surveyed ($X$) 1 15 200 2 22 150 3 18 180 4 30 220 5 25 300

 a. 21490.9 b. 210 c. 2.15 d. 215

Solution:

Weighted mean of the percentage of people working in night shifts is X = ∑wX / ∑w
[Formula.]

wX = 15 / 100 × 200 + 22 / 100 × 150 + 18 / 100 × 180 + 30 / 100 × 220 + 25 / 100 ×300 = 30 + 33 + 32.4 + 66 + 75

w = 15 / 100 + 22 / 100 + 18 / 100 + 30 / 100 + 25 / 100 = 110 / 100

∑wX / ∑w = (30+33+32.4+66+75)×100110
[Substitute the values and simplify.]

= 2.15 × 100 = 215

So, the weighted mean is 215.

Correct answer : (4)
8.
Which of the following statement(s) is/are true?
I. The mode for the traffic fatalities for 10 selected states in a specific year (1488, 1586, 4192, 1031, 645, 74, 158, 875, 3360, 325) is zero.
II. The mode is not always unique for a data set.
III. The midrange of 2, 3, 6, 8, 4, 1 is 3.5.
IV. The midrange and median for the data in statement III are same.
 a. III only b. II and III only c. II only d. II, III and IV only

Solution:

1488, 1586, 4192, 1031, 645, 74, 158, 875, 3360, 325

Each value occurs only once, so there is no mode for the above data.

But writing the mode for the dataset as zero is incorrect. So, statement I is false.

The mode is not always unique. A dataset can have more than one mode, or mode may not exist for a dataset. So, statement II is true.

Mid range of the data 2, 3, 6, 8, 4, 1 is 1+82 = 9 / 2 = 4.5
[Use Mid range = lowest value+highest value2.]

So, statement III is false.

1, 2, 3, 4, 6, 8
[Arrange the data of statement III in order.]

Median = 3+42 = 7 / 2 = 3.5
[Median is average of middle values in the data set.]

The median for the above data is 3.5 and midrange is 4.5. Both are not same.

Statement IV is false.

So, only statement II is correct.

Correct answer : (3)
9.
Describe which measure of central tendency - mean, median, or mode was probably used in each situation. Answer in order.
I. One half of the gross domestic product (GDP) of country A comes from more than 42.5% of its citizens.
II. Computation of per-capita income in several countries.
III. A person on an average eats sea food once a week.
IV. Average sales of a company for various years.
 a. mean, mode, mode, mode b. mode, mode, mean, median c. median, mean, mode, mean d. mean, mean, mean, median

Solution:

Statement I: One half of the gross domestic product (GDP) of country A comes from more than 42.5% of its citizens and one half comes from less than 42.5% of its citizens. So, median was used.
[Since median is the half way point in a data set.]

Statement II: The mean for a data set is unique. So, mean was the correct measure of central tendency to compute the per-capita income in several countries.

Statement III: Mode can be used when the data are nominal such as preferences, gender etc. So, mode was probably used.

Statement IV: Mean was used to know the average sales of a company for various years.

Correct answer : (3)
10.
Which of the following statement(s) is/are true?
I. The population mean is represented by the Roman letter $\stackrel{‾}{X}$. The sample mean is represented by the Greek letter $\mu$.
II. The measures that determine the spread of the data values are called measures of dispersion.
III. The ages of middle school children whose median age is 10 years are all below 10 years.
IV. The mean of the salaries of all the employees working in a small company is represented by $\mu$.
 a. II and IV only b. I and III only c. IV only d. all are correct

Solution:

Greek letters are used to denote parameters. The population mean is represented by the Greek letter μ.

Roman letters are used to denote statistics. The sample mean is represented by the Roman letter X.

So, statement I is false.

To know, whether the data values cluster around the mean or are they spread more evenly through out the distribution, the measures used are called the measures of dispersion or measures of variation.

So, statement II is true.

Median is the half way point in the data set. Median age is 10 years means the age of one half of middle school children is less than 10 years and the age of one half of children is more than 10 years.

So, statement III is false.

The mean of the salaries obtained of all the employees in a small company (population) is represented by the Greek letter μ.

So, statement IV is true.

Correct answer : (1)

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