Business Statistics (1430)
(a)Q. 1 Differentiate between populations
and samples, and describe some advantages of samples over populations. (10)
(b) Why a frequency distribution is
constructed? Explain various steps involved in the construction of a frequency
distribution. (10)
(a) Differentiating between populations and samples and advantages of samples over populations:
Population:
In
statistics, a population refers to the entire group or set of individuals,
items, or data that share a common characteristic and are of interest to a
researcher. For example, if a researcher wants to study the average height of
all adult males in a country, the population would include all adult males in
that country. Populations can be finite, like the number of students in a
school, or infinite, like the possible outcomes of rolling a dice an infinite
number of times.
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Sample:
A
sample, on the other hand, is a subset or a smaller representative group taken
from the entire population. Researchers use samples to draw inferences and make
predictions about the entire population without having to study every
individual in it. In the height example, instead of measuring the height of
every adult male in the country, the researcher can select a smaller group of
adult males, measure their heights, and use that information to make inferences
about the entire population.
Advantages of samples over populations:
1. Cost and Time Efficiency:
Studying an entire population can be time-consuming, expensive, and sometimes
impractical. By using a sample, researchers can collect the necessary data
efficiently and at a lower cost.
2. Feasibility: In
some cases, it may not be feasible to access or gather data from an entire
population due to geographical, logistical, or other constraints. A sample
allows researchers to work with more manageable and accessible data.
3. Reduced Bias: When
dealing with large populations, collecting data from every individual can
introduce errors and biases. A carefully selected and representative sample can
help minimize such biases, leading to more accurate results.
4. Ethical Considerations: In
situations where studying the entire population could cause harm or discomfort
to individuals, using a sample that provides comparable insights can be a more
ethical approach.
5. Extrapolation: By
analyzing a representative sample, researchers can make valid inferences and
generalize their findings to the entire population, which is often the goal of
statistical research.
6. Testing and Piloting: Before
conducting large-scale research or experiments, researchers often pilot their
studies on a sample to identify potential flaws or areas for improvement in
their methodologies.
7. Manageable Data Analysis: Dealing
with data from an entire population can be overwhelming. Working with a sample
allows researchers to perform more focused and manageable data analysis.
(b) Explanation of constructing a frequency
distribution and steps involved:
Frequency Distribution:
A
frequency distribution is a tabular or graphical representation that shows the
number of occurrences (frequency) of each distinct value or category in a
dataset. It helps to organize and summarize data, making it easier to
understand patterns, trends, and distributions within the dataset.
Steps involved in constructing a frequency
distribution:
1. Data Collection:
The
first step in constructing a frequency distribution is to collect the data.
This could be done through surveys, experiments, observations, or any other
data collection method appropriate for the study.
2. Data Organization:
Once
the data is collected, it needs to be organized in a systematic manner. Arrange
the data in ascending or descending order, depending on the nature of the data,
to facilitate the construction of the frequency distribution.
3. Determine the Number of Classes
(Categories):
The next
step is to decide how many classes or categories will be used to group the
data. The number of classes should be sufficient to capture the variation in
the data but not too large to lose the essential information. The choice of the
number of classes can be based on various methods, such as the Sturges formula
or the square root rule.
4. Calculate the Range:
Find
the range of the data, which is the difference between the maximum and minimum
values. The range helps in determining the width of each class interval.
5. Calculate the Class Width:
To
calculate the class width, divide the range by the number of classes. Round up
to the nearest convenient number to ensure that the classes are of equal width.
6. Create Class Intervals:
Using
the class width, create non-overlapping intervals or bins that cover the entire
range of the data. The class intervals should be continuous and should not
leave any gaps.
7. Tally the Data:
Go
through the data and place each observation into its corresponding class
interval. Keep a tally or count of the number of data points in each interval.
8. Calculate Frequencies:
Convert
the tallies to actual frequencies, which represent the number of occurrences of
data points in each class interval.
9. Optional: Construct Cumulative Frequency
Distribution:
If
desired, a cumulative frequency distribution can be constructed. It shows the
total number of data points that fall within or below each class interval.
10. Optional: Create a Frequency Table or
Graph:
Present
the frequency distribution in a tabular form or visually represent it using
graphs like histograms, bar charts, or frequency polygons. A graph makes it
easier to visualize the data distribution and identify any patterns or
outliers.
In
conclusion, constructing a frequency distribution is an essential step in
organizing and summarizing data, making it easier to extract meaningful
insights and draw conclusions from the dataset. By following the steps
mentioned above, researchers can effectively analyze and present their data in
a clear and concise manner.
Q. 2 A
certain Transportation Commission is concerned about the speed motorists are
deriving on a section of the main highway. Here are the speeds of 45 motorists:
15 32 45 46 42 39 68 47 18 31 48 49
56 52 39 48 69 61 44 42 38 52 55 58
62 58 48 56 58 48 47 52 37 64 29 55
38 29 62 49 69 18 61 55
Use
these data to construct relative frequency distributions using 5 equal
intervals and II equal intervals. The Department of transportation (DOT)
reports that no more than 10 percent of the motorists exceed 55 mph.
Do the motorists follow the DOT's report
about driving pattern? Which distribution did you use to answer this part? (10)
(b) The
DOT has determined that the safest speed for this highway is more than 36 but
less than 59 mph. What percent of the motorists drive within this range? Which
distribution did you use to answer this part?
(10)
To
construct the relative frequency distributions using 5 equal intervals and 11
equal intervals, we first need to organize the given data in ascending order.
Then, we will calculate the frequency of each interval and the relative
frequency (percentage) of motorists falling within each interval. Once the
distributions are constructed, we can answer the questions regarding the
driving patterns and the percentage of motorists driving within the specified
range.
Given data:
15 32
45 46 42 39 68 47 18 31 48 49
56 52
39 48 69 61 44 42 38 52 55 58
62 58
48 56 58 48 47 52 37 64 29 55
38 29
62 49 69 18 61 55
(a) Constructing relative frequency
distributions using 5 equal intervals and 11 equal intervals:
Step 1: Organize the data in
ascending order:
15 18
18 29 29 31 32 37 38 38 39 39 42 44 45 46 47 47 48 48 48 49
49 52
52 52 55 55 55 56 56 58 58 58 61 61 62 62 64 68 69 69
Step 2: Determine the range and
the width of intervals:
Range
= Maximum value - Minimum value
Range
= 69 - 15 = 54
Width of intervals for 5 equal intervals:
Interval
width = Range / Number of intervals
Interval
width = 54 / 5 = 10.8 (Approximately 11)
Width of intervals for 11 equal intervals:
Interval
width = Range / Number of intervals
Interval
width = 54 / 11 ≈ 4.91 (Approximately 5)
Step 3: Create intervals for 5
equal intervals:
Interval 1: 15 -
25 (11 motorists)
Interval 2: 26 -
36 (4 motorists)
Interval 3: 37 -
47 (7 motorists)
Interval 4: 48 -
58 (16 motorists)
Interval 5: 59 -
69 (7 motorists)
Step 4: Create intervals for 11
equal intervals:
Interval 1: 15 -
20 (2 motorists)
Interval 2: 21 -
26 (2 motorists)
Interval 3: 27 -
32 (2 motorists)
Interval 4: 33 -
38 (5 motorists)
Interval 5: 39 -
44 (6 motorists)
Interval 6: 45 -
50 (9 motorists)
Interval 7: 51 -
56 (11 motorists)
Interval 8: 57 -
62 (7 motorists)
Interval 9: 63 -
68 (2 motorists)
Interval 10: 69 -
74 (2 motorists)
Step 5: Calculate the frequency
and relative frequency (percentage) for each interval:
For 5 equal intervals:
Interval 1: 11
motorists (Frequency) - 24.44% (Relative frequency)
Interval 2: 4
motorists (Frequency) - 8.88% (Relative frequency)
Interval 3: 7
motorists (Frequency) - 15.55% (Relative frequency)
Interval 4: 16
motorists (Frequency) - 35.55% (Relative frequency)
Interval 5: 7
motorists (Frequency) - 15.55% (Relative frequency)
For 11 equal intervals:
Interval 1: 2
motorists (Frequency) - 4.44% (Relative frequency)
Interval 2: 2
motorists (Frequency) - 4.44% (Relative frequency)
Interval 3: 2
motorists (Frequency) - 4.44% (Relative frequency)
Interval 4: 5
motorists (Frequency) - 11.11% (Relative frequency)
Interval 5: 6
motorists (Frequency) - 13.33% (Relative frequency)
Interval 6: 9
motorists (Frequency) - 20.00% (Relative frequency)
Interval 7: 11
motorists (Frequency) - 24.44% (Relative frequency)
Interval 8: 7
motorists (Frequency) - 15.55% (Relative frequency)
Interval 9: 2
motorists (Frequency) - 4.44% (Relative frequency)
Interval 10: 2
motorists (Frequency) - 4.44% (Relative frequency)
(b) Analyzing the results and answering
the questions:
(a) Do the motorists follow the DOT's
report about the driving pattern? Which distribution did you use to answer this
part?
To
answer this part, we need to check if no more than 10 percent of the motorists
exceed 55 mph. From both distributions, we can see that the percentage of
motorists driving above 55 mph is more than 10 percent in both cases.
In the
5 equal intervals distribution, there are 7 motorists (15.55%) in the last
interval (59-69), which exceeds the 10 percent threshold.
In the
11 equal intervals distribution, there are 2 motorists (4.44%) in the last
interval (69-74), which also exceeds the 10 percent threshold.
Based
on the data, it appears that the motorists do not fully follow the DOT's report
about the driving pattern, as more than 10 percent of them are driving above 55
mph.
(b) The DOT has determined that the safest
speed for this highway is more than 36 but less than 59 mph. What percent of
the motorists drive within this range? Which distribution did you use to answer
this part?
To
answer this part, we need to calculate the percentage of motorists who fall
within the range of more than 36 but less than 59 mph. We can use either
distribution to calculate this percentage.
In the
5 equal intervals distribution, the motorists falling within the range (37-58)
are in Interval 3 and Interval 4:
Interval 3: 7
motorists (15.55%)
Interval 4: 16
motorists (35.55%)
Total
percentage of motorists driving within the range: 15.55% + 35.55% = 51.10%
In the
11 equal intervals distribution, the motorists falling within the range (39-58)
are in Interval 5, Interval 6, Interval 7, and Interval 8:
Interval 5: 6
motorists (13.33%)
Interval 6: 9
motorists (20.00%)
Interval 7: 11
motorists (24.44%)
Interval 8: 7
motorists (15.55%)
Total
percentage of motorists driving within the range: 13.33% + 20.00% + 24.44% +
15.55% = 73.32%
Based
on the data from both distributions, a higher percentage of motorists (73.32%)
drive within the range specified by the DOT (more than 36 but less than 59 mph)
in the 11 equal intervals distribution. Therefore, the 11 equal intervals
distribution provides a more accurate representation of the percentage of
motorists driving within the specified range on the highway.
In
conclusion, the relative frequency distributions using 5 equal intervals and 11
equal intervals show that a significant portion of motorists exceed the speed
limit of 55 mph, as reported
by the
DOT. However, a higher percentage of motorists drive within the safe speed
range of more than 36 but less than 59 mph, especially when using the 11 equal
intervals distribution. It is important for the Transportation Commission to
consider these findings to address speeding concerns and promote safer driving
habits on the highway.
Q. 3 (a)
Differentiate between simple
histogram and relative frequency histogram. (10)
(b) Here is a frequency distribution of length of phone calls made
by 175 people during a Labor Day weekend. Construct histogram and frequency
polygon for these data. (10)
|
Length in Minutes |
1–7 |
8–14 |
15–21 |
22–28 |
29–35 |
39–42 |
43–49 |
50–56 |
|
Frequency |
45 |
32 |
34 |
22 |
16 |
12 |
9 |
5 |
(a) Differentiating between a simple
histogram and a relative frequency histogram:
1. Simple Histogram:
A
simple histogram is a graphical representation of the distribution of numerical
data. It consists of bars or rectangles, where the width of each bar represents
a class interval and the height of the bar corresponds to the frequency or
count of data points falling within that interval. The data is grouped into
intervals or bins, and the bars are placed above each interval on the
horizontal axis. Simple histograms are used to visualize the distribution and
frequency of data, helping to identify patterns, trends, and outliers.
2. Relative Frequency Histogram:
A
relative frequency histogram is similar to a simple histogram, but instead of
showing the actual frequencies or counts, it represents the relative
frequencies or proportions of each class interval. The relative frequency of a
class interval is calculated by dividing the frequency of that interval by the
total number of data points. The height of each bar in a relative frequency
histogram represents the proportion or percentage of data points falling within
that interval. Relative frequency histograms are particularly useful when
comparing datasets with different sample sizes or when analyzing data with
varying total counts.
Main differences:
-
Scale: In a simple histogram, the vertical axis represents the actual
frequencies or counts of data points, while in a relative frequency histogram,
it represents the relative frequencies or proportions.
- Interpretation:
Simple histograms are useful for understanding the distribution and frequency
of data in absolute terms, while relative frequency histograms allow for the
comparison of proportions across different datasets or sample sizes.
- Y-axis units: The
y-axis of a simple histogram shows the actual counts, whereas the y-axis of a
relative frequency histogram displays proportions or percentages.
- Total area: The
total area under the bars in a simple histogram represents the total number of
data points, while in a relative frequency histogram, the total area under the
bars adds up to 1 (or 100% if represented as percentages).
- Use cases:
Simple histograms are commonly used when the focus is on the actual counts of
data points, while relative frequency histograms are more suitable when the
emphasis is on understanding the proportion of data points within each interval
relative to the total data.
(b) Constructing a histogram and frequency
polygon for the given frequency distribution:
To construct a histogram and frequency
polygon, we will follow these steps:
Step 1: Organize the data into
a table with two columns - Length in Minutes (class intervals) and Frequency.
Length
in Minutes | Frequency
-----------------|----------
1–7 | 45
8–14 | 32
15–21 | 34
22–28 | 22
29–35 | 16
39–42 | 12
43–49 | 9
50–56 | 5
Step 2: Determine the range and
the width of intervals:
The
range is the difference between the maximum and minimum values of the class
intervals.
Range
= (56 - 1) = 55
Next,
we need to determine the width of each interval. For this, we can choose any
convenient number, but it should be consistent with the data and not too wide
or too narrow. Let's choose a width of 7.
Step 3: Create the histogram:
Now,
we will construct the histogram using the frequency distribution data:
Length
in Minutes | Frequency | Histogram
-----------------|-----------|----------
1–7 | 45 | ***********
8–14 | 32 | ********
15–21 | 34 | *********
22–28 | 22 | ******
29–35 | 16 | ****
39–42 | 12 | ***
43–49 | 9 | **
50–56 | 5 | *
In the
histogram, each bar's length is determined by the frequency of the
corresponding class interval. The number of asterisks (*) represents the
frequency in each interval, and the width of each interval is 7 minutes.
Step 4: Create the frequency
polygon:
A
frequency polygon is a line graph that represents the frequencies of the class
intervals. It is constructed by plotting points at the midpoint of each
interval and then connecting the points with straight lines.
To
construct the frequency polygon, we need to find the midpoints of each class
interval:
Midpoint
= (Lower Limit + Upper Limit) / 2
Midpoint for each interval:
1–7: (1 + 7) / 2 = 4
8–14: (8 + 14) / 2 = 11
15–21: (15 + 21) / 2 = 18
22–28: (22 + 28) / 2 = 25
29–35: (29 + 35) / 2 = 32
39–42: (39 + 42) / 2 = 40.5
43–49: (43 + 49) / 2 = 46
50–56: (50 + 56) / 2 = 53
Now, we can plot the frequency polygon:
Frequency Polygon:
(4,
45) - (11, 32) - (18, 34) - (25, 22) - (32, 16) - (40.5, 12) - (46, 9) - (53,
5)
In the
frequency polygon, each point represents the midpoint of the corresponding
class interval, and the vertical position of the point represents the
frequency. The points are then connected with straight lines.
By
constructing the histogram and frequency polygon for the given frequency
distribution, we have visually represented the distribution of phone call
lengths during the Labor Day weekend. These graphical representations help to
better understand the patterns and frequencies of phone call lengths made by
175 people during that time.
Q. 4 The
administrator of a hospital surveyed the number of days 200 randomly chosen
patients stayed in the hospital following an operation. The data are.
|
Hospital stay in Days |
1–3 |
4–6 |
7–9 |
10–12 |
13–15 |
16–18 |
19–21 |
22–24 |
|
Frequency |
18 |
90 |
44 |
21 |
9 |
9 |
4 |
5 |
(a) Calculate mean, standard deviation and
coefficient of variation. (10)
(b) According to Chebyshev's theorem, how
many stays should be between 0 and 17 days? How many are actually in that
interval? (05)
(c) Because the distribution is roughly
bell-shaped, how many stays can we expect between 0 and 17 days? (05)
(a) Calculating mean, standard deviation,
and coefficient of variation:
Step 1: Set up the data in a
table with two columns - Hospital Stay in Days (class intervals) and Frequency.
Hospital
Stay in Days | Frequency
---------------------|----------
1–3 | 18
4–6 | 90
7–9 | 44
10–12 | 21
13–15 | 9
16–18 | 9
19–21 | 4
22–24 | 5
Step 2: Find the midpoint of
each class interval:
Midpoint
= (Lower Limit + Upper Limit) / 2
Midpoint for each interval:
1–3: (1 + 3) / 2 = 2
4–6: (4 + 6) / 2 = 5
7–9: (7 + 9) / 2 = 8
10–12: (10 + 12) / 2 = 11
13–15: (13 + 15) / 2 = 14
16–18: (16 + 18) / 2 = 17
19–21: (19 + 21) / 2 = 20
22–24:
(22 + 24) / 2 = 23
Step 3: Calculate the weighted
mean:
Weighted
Mean = Σ (Midpoint × Frequency) / Σ Frequency
Weighted
Mean = (2×18 + 5×90 + 8×44 + 11×21 + 14×9 + 17×9 + 20×4 + 23×5) / (18 + 90 + 44
+ 21 + 9 + 9 + 4 + 5)
Weighted
Mean = (36 + 450 + 352 + 231 + 126 + 153 + 80 + 115) / 200
Weighted
Mean = 1563 / 200
Weighted
Mean = 7.815
Step 4: Calculate the standard
deviation:
To
calculate the standard deviation, we first need to calculate the variance.
Variance
= Σ [(Midpoint - Mean)² × Frequency] / Σ Frequency
Variance
= [(2 - 7.815)² × 18 + (5 - 7.815)² × 90 + (8 - 7.815)² × 44 + (11 - 7.815)² ×
21 + (14 - 7.815)² × 9 + (17 - 7.815)² × 9 + (20 - 7.815)² × 4 + (23 - 7.815)²
× 5] / 200
Variance
= [(-5.815)² × 18 + (-2.815)² × 90 + (0.185)² × 44 + (3.185)² × 21 + (6.185)² ×
9 + (9.185)² × 9 + (12.185)² × 4 + (15.185)² × 5] / 200
Variance
= [33.841 × 18 + 7.928 × 90 + 0.034 × 44 + 10.161 × 21 + 38.292 × 9 + 84.508 ×
9 + 148.741 × 4 + 229.807 × 5] / 200
Variance
= (608.538 + 713.52 + 1.496 + 213.378 + 344.628 + 760.572 + 594.964 + 1149.035)
/ 200
Variance
= 4384.151 / 200
Variance
= 21.920755
Standard
Deviation = √Variance
Standard
Deviation = √21.920755
Standard
Deviation ≈ 4.678
Step 5: Calculate the
coefficient of variation (CV):
Coefficient
of Variation (CV) = (Standard Deviation / Mean) × 100
CV =
(4.678 / 7.815) × 100
CV ≈
59.86%
(b) According to Chebyshev's theorem, how
many stays should be between 0 and 17 days? How many are actually in that
interval?
Chebyshev's
theorem states that for any distribution (regardless of its shape), the
proportion of data points that fall within k standard deviations from the mean
is at least (1 - 1/k²), where k is a positive constant greater than 1.
In our case, we already calculated the
mean and standard deviation:
Mean =
7.815
Standard
Deviation = 4.678
To
find the number of stays that should be between 0 and 17 days according to
Chebyshev's theorem, we will use k = 2 (since k must be greater than 1).
Proportion
within 2 standard deviations from the mean = 1 - 1/2² = 1 - 1/4 = 3/4
Number
of stays within 2 standard deviations from the mean = Proportion × Total number
of data points
Number
of stays within 2 standard deviations = (3/4) × 200 = 150
Now, let's see how many stays are actually
in that interval (0-17 days):
From the given frequency distribution, we
have:
1-3 days: 18
stays
4-6 days: 90
stays
7-9 days: 44
stays
10-12 days: 21
stays
13-15 days: 9
stays
16-18 days: 9
stays
Total
stays in the 0-17 days interval = 18 + 90 + 44 + 21 + 9 + 9 = 191
So,
according to Chebyshev's theorem, at least 150 stays should be between 0 and 17
days. However, in the given data, there are actually 191 stays in that
interval.
(c) Because the distribution is roughly
bell-shaped, how many stays can we expect between 0 and 17 days?
Since
the distribution is roughly bell-shaped, we can expect a higher concentration
of data points around the mean, and a symmetrical distribution. In a normal
distribution (bell-shaped), the majority of data points are clustered around
the mean, with a decreasing number of data points as we move away from the
mean.
Based
on Chebyshev's theorem, we can expect at least 150 stays (75% of the data) to
be within 2 standard deviations from the mean. However, since the distribution
is roughly bell-shaped, we can expect an even higher percentage of stays to be
within that range.
Q. 5 (a)
Differentiate between the following:
(i) Type I and Type II errors, (ii) Two-tailed and one tailed tests of
hypotheses and (Hi) Hypotheses testing of means when the population standard
deviation is known and not known. (10)
(b) For a sample of 60 women taken from
population of over 5000 enrolled in a weight-reducing program
(a) Calculating mean, standard deviation,
and coefficient of variation:
Step 1: Set up the data in a
table with two columns - Hospital Stay in Days (class intervals) and Frequency.
Hospital
Stay in Days | Frequency
---------------------|----------
1–3 | 18
4–6 | 90
7–9 | 44
10–12 | 21
13–15 | 9
16–18 | 9
19–21 | 4
22–24 | 5
Step 2: Find the midpoint of
each class interval:
Midpoint
= (Lower Limit + Upper Limit) / 2
Midpoint for each interval:
1–3: (1 + 3) / 2 = 2
4–6: (4 + 6) / 2 = 5
7–9: (7 + 9) / 2 = 8
10–12: (10 + 12) / 2 = 11
13–15: (13 + 15) / 2 = 14
16–18: (16 + 18) / 2 = 17
19–21: (19 + 21) / 2 = 20
22–24: (22 + 24) / 2 = 23
Step 3: Calculate the weighted
mean:
Weighted
Mean = Σ (Midpoint × Frequency) / Σ Frequency
Weighted
Mean = (2×18 + 5×90 + 8×44 + 11×21 + 14×9 + 17×9 + 20×4 + 23×5) / (18 + 90 + 44
+ 21 + 9 + 9 + 4 + 5)
Weighted
Mean = (36 + 450 + 352 + 231 + 126 + 153 + 80 + 115) / 200
Weighted
Mean = 1563 / 200
Weighted
Mean = 7.815
Step 4: Calculate the standard
deviation:
To
calculate the standard deviation, we first need to calculate the variance.
Variance
= Σ [(Midpoint - Mean)² × Frequency] / Σ Frequency
Variance
= [(2 - 7.815)² × 18 + (5 - 7.815)² × 90 + (8 - 7.815)² × 44 + (11 - 7.815)² ×
21 + (14 - 7.815)² × 9 + (17 - 7.815)² × 9 + (20 - 7.815)² × 4 + (23 - 7.815)²
× 5] / 200
Variance
= [(-5.815)² × 18 + (-2.815)² × 90 + (0.185)² × 44 + (3.185)² × 21 + (6.185)² ×
9 + (9.185)² × 9 + (12.185)² × 4 + (15.185)² × 5] / 200
Variance
= [33.841 × 18 + 7.928 × 90 + 0.034 × 44 + 10.161 × 21 + 38.292 × 9 + 84.508 ×
9 + 148.741 × 4 + 229.807 × 5] / 200
Variance
= (608.538 + 713.52 + 1.496 + 213.378 + 344.628 + 760.572 + 594.964 + 1149.035)
/ 200
Variance
= 4384.151 / 200
Variance
= 21.920755
Standard
Deviation = √Variance
Standard
Deviation = √21.920755
Standard
Deviation ≈ 4.678
Step 5: Calculate the
coefficient of variation (CV):
Coefficient
of Variation (CV) = (Standard Deviation / Mean) × 100
CV =
(4.678 / 7.815) × 100
CV ≈
59.86%
(b) According to Chebyshev's theorem, how
many stays should be between 0 and 17 days? How many are actually in that
interval?
Chebyshev's
theorem states that for any distribution (regardless of its shape), the
proportion of data points that fall within k standard deviations from the mean
is at least (1 - 1/k²), where k is a positive constant greater than 1.
In our
case, we already calculated the mean and standard deviation:
Mean =
7.815
Standard
Deviation = 4.678
To
find the number of stays that should be between 0 and 17 days according to
Chebyshev's theorem, we will use k = 2 (since k must be greater than 1).
Proportion
within 2 standard deviations from the mean = 1 - 1/2² = 1 - 1/4 = 3/4
Number
of stays within 2 standard deviations from the mean = Proportion × Total number
of data points
Number
of stays within 2 standard deviations = (3/4) × 200 = 150
Now,
let's see how many stays are actually in that interval (0-17 days):
From
the given frequency distribution, we have:
1-3 days: 18
stays
4-6 days: 90
stays
7-9 days: 44
stays
10-12 days: 21
stays
13-15 days: 9
stays
16-18 days: 9
stays
Total
stays in the 0-17 days interval = 18 + 90 + 44 + 21 + 9 + 9 = 191
So,
according to Chebyshev's theorem, at least 150 stays should be between 0 and 17
days. However, in the given data, there are actually 191 stays in that
interval.
(c) Because the distribution is roughly
bell-shaped, how many stays can we expect between 0 and 17 days?
Since
the distribution is roughly bell-shaped, we can expect a higher concentration
of data points around the mean, and a symmetrical distribution. In a normal
distribution (bell-shaped), the majority of data points are clustered around
the mean, with a decreasing number of data points as we move away from the
mean.
Based
on Chebyshev's theorem, we can expect at least 150 stays (75% of the data) to
be within 2 standard deviations from the mean. However, since the distribution
is roughly bell-shaped, we can expect an even higher percentage of stays to be
within that range. Dear
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