Solution
Atul answered on
May 04 2023
Question 1
Groups Frequencies
300 to 307 12
307 to 314 18
314 to 321 44
321 to 328 88
328 to 335 86
335 to 342 41
342 to 349 15
349 to 356 9
The lifetimes (in units of 106 seconds) of certain satellite components are shown in the
frequency distribution given in ‘Dataset1’.
1. Draw a frequency polygon, histogram and cumulative frequency polygon for the
data.
To draw the frequency polygon, we first need to calculate the midpoints of each group:
Groups Frequencies Midpoints
300 to 307 12 303.5
307 to 314 18 310.5
314 to 321 44 317.5
321 to 328 88 324.5
328 to 335 86 331.5
335 to 342 41 338.5
342 to 349 15 345.5
349 to 356 9 352.5
Histogram
Finally, to draw the cumulative frequency polygon, we need to calculate the cumulative
frequencies:
Groups Midpoints Frequencies
Cumulative
Frequencies
300-307 303.5 12 12
307-314 310.5 18 30
314-321 317.5 44 74
321-328 324.5 88 162
328-335 331.5 86 248
335-342 338.5 41 289
342-349 345.5 15 304
349-356 352.5 9 313
2. Calculate the frequency mean, the frequency standard deviation, the median and the
first and third quartiles for this grouped data.
To calculate the mean, we need to use the midpoint of each group and the frequency of each
group:
Midpoint Frequency Midpoint * Frequency
----------------------------------------------
303.5 12 3,642
310.5 18 5,589
317.5 44 13,970
324.5 88 28,556
331.5 86 28,521
338.5 41 13,895
345.5 15 5,183
352.5 9 3,173
----------------------------------------------
102,509
The total frequency is 323, so the frequency mean is:
Mean = 102,509 / 323 ≈ 317.44
To calculate the frequency standard deviation, we need to find the variance first. We can use
the formula:
Variance = (Σ (f * (x - mean)²)) / (N - 1)
where f is the frequency, x is the midpoint, mean is the frequency mean, and N is the total
frequency.
Midpoint Frequency x - mean (x - mean)² f * (x - mean)²
--------------------------------------------------------------------
303.5 12 -13.94 193.8756 2326.5072
310.5 18 -6.94 48.1636 866.9450
317.5 44 0.56 0.3136 13.7904
324.5 88 7.56 57.1536 5028.9792
331.5 86 14.56 211.9936 18223.0784
338.5 41 21.56 464.6736 19029.7856
345.5 15 28.56 817.5936 12263.9040
352.5 9 35.56 1262.0736 11358.6624
--------------------------------------------------------------------
Σ 66487.6624
Variance = 66487.6624 / (323 - 1) ≈ 207.8
Standard deviation = √207.8 ≈ 14.42
To calculate the median, we need to find the frequency cumulative distribution function
(CDF)
Midpoint Frequency Cumulative frequency
----------------------------------------------
303.5 12 12
310.5 18 30
317.5 44 74
324.5 88 162
331.5 86 248
338.5 41 289
345.5 15 304
352.5 9 313
The total frequency is 323, so the median co
esponds to the midpoint that has a cumulative
frequency of 161.5, which lies between the fourth and fifth groups. To estimate the median,
we can use the formula:
Median = L + ((N / 2 - CF(L-1)) / f) * w
where L is the lower limit of the group that contains the median, N is the total frequency,
CF(L-1) is the cumulative frequency up to the previous group, f is the frequency of the group
that contains the median, and w is the width of the group.
In this case, we have :
L = 321
N = 323
CF(L-1) = 30
f = 88
w = 7
Median = 321 + ((323 / 2 - 30) / 88) * 7 ≈ 326.43
To find the quartile boundaries, we can use the formula:
Q(n) = L + ((n / 4 * N) - CF(L-1)) / f * w
where n is the quartile number (1 for the first quartile, 3 for the third quartile), and the other
variables have the same meaning as before.
For the first quartile, we have:
n = 1
Q(1) = L + ((1 / 4 * 323) - CF(L-1)) / f * w
= 321 + ((0.25 * 323) - 30) / 88 * 7
= 313.73
For third Quartile
n = 3
Q(3) = L + ((3 / 4 * 323) - CF(L-1)) / f * w
= 331.5 + ((0.75 * 323) - 248) / 86 * 7
= 340.34
So, the estimated first quartile is 313.73, and the estimated third quartile is 340.34.
3. Compare the median and the mean and state what this indicates about the
distribution. Comment on how the answer to this question relates to your frequency
polygon and histogram.
Comparison of Median and Mean
The median of the data set is 326.43, and the mean is 317.44. The fact that the median is
slightly larger than the mean indicates that the distribution is slightly skewed to the right.
This is consistent with what we see in the frequency polygon and histogram, where there are
more values on the right side of the distribution.
4. Explain the logic behind the equations for the mean and standard deviation for
grouped data, starting from the original equations for a simple list of data values. (This
does not just mean ’explain how the equations are used’.)
The equations for the mean and standard deviation for grouped data are modifications of the
equations for the mean and standard deviation for a simple list of data values. The main
difference is that the grouped data is divided into intervals, and the frequency of each interval
is used to determine the weight of each interval in the calculation of the mean and standard
deviation.
For the mean, the equation for grouped data is:
mean = Σ (midpoint * frequency) / Σ frequency
where midpoint is the midpoint of each interval, and frequency is the frequency of each
interval. The numerator represents the sum of the products of the midpoint and frequency of
each interval, while the denominator represents the total frequency of all intervals. This
equation is used to calculate the weighted average of the midpoints of the intervals, where the
weight of each interval is its frequency.
For the standard deviation, the equation for grouped data is:
standard deviation = sqrt(Σ [(x - mean)^2 * frequency] / (Σ frequency - 1))
where x is the midpoint of each interval, mean is the mean of the data set, and frequency is
the frequency of each interval. The numerator represents the sum of the products of the
squared differences between the midpoint and the mean and the frequency of each interval,
while the denominator represents the total frequency of all intervals minus one. This equation
is used to calculate the weighted average of the squared deviations of the midpoints from the
mean, where the weight of each interval is its frequency.
The modification of the equations is necessary because grouped data provides less
information about the individual data points than a simple list of values. The midpoint of each
interval is used to represent all the data points within the interval, and the frequency of each
interval is used to determine the weight of each interval in the calculation of the mean and
standard deviation.
5.Ca
y out an appropriate statistical test to determine whether the data is normally
distributed.
We can use the following method Anderson-Darling test for the given grouped data, we first
need to calculate the expected frequencies for a normal distribution with the same mean and
standard deviation as the data. We can use the following formula to calculate the expected
frequency for an interval:
Expected frequency = (Φ(upper bound) - Φ(lower bound)) * N
where Φ() is the cumulative distribution function of the standard normal distribution, upper
ound and lower bound are the upper and lower bounds of the interval, and N is the total
sample size.
Using the given data, we can calculate the sample mean and sample standard deviation as
follows:
mean = (300+307)*12/2 +...