Answer To: Statistics and Probability
Atul answered on May 06 2023
Question 1
Groups Frequencies
300 to 305 6
305 to 310 10
310 to 315 35
315 to 320 81
320 to 325 82
325 to 330 38
330 to 335 14
335 to 340 10
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:
Intervals Frequencies Midpoint Cumulative Frequency
300-305 6 302.5 6
305-310 10 307.5 16
310-315 35 312.5 51
315-320 81 317.5 132
320-325 82 322.5 214
325-330 38 327.5 252
330-335 14 332.5 266
335-340 10 337.5 276
Histogram
Finally, to draw the cumulative frequency polygon, we need to calculate the cumulative
frequencies:
Intervals Frequencies Midpoint
Cumulative
Frequency
300-305 6 302.5 6
305-310 10 307.5 16
310-315 35 312.5 51
315-320 81 317.5 132
320-325 82 322.5 214
325-330 38 327.5 252
330-335 14 332.5 266
335-340 10 337.5 276
To calculate the frequency mean, we need to first calculate the midpoint of each interval, then
multiply each midpoint by its corresponding frequency, sum up the results, and finally divide
by the total frequency.
Intervals Frequencies Midpoint
300-305 6 302.5
305-310 10 307.5
310-315 35 312.5
315-320 81 317.5
320-325 82 322.5
325-330 38 327.5
330-335 14 332.5
335-340 10 337.5
Frequency Mean = (6*302.5 + 10*307.5 + 35*312.5 + 81*317.5 + 82*322.5 + 38*327.5 +
14*332.5 + 10*337.5) / (6+10+35+81+82+38+14+10) = 320.7
The frequency standard deviation can be calculated using the following formula:
σ = sqrt[(Σ(f(x) - mean)^2) / n]
where f(x) is the frequency of each interval, mean is the frequency mean we just calculated,
and n is the total frequency.
f(x) midpoint deviation (deviation)^2 f(x)*(deviation)^2
6 302.5 -18.2 331.24 1987.44
10 307.5 -13.2 174.24 1742.4
35 312.5 -8.2 67.24 2353.4
81 317.5 -2.2 4.84 392.04
82 322.5 2.8 7.84 642.88
38 327.5 7.8 60.84 2312.92
14 332.5 12.8 163.84 2293.76
10 337.5 17.8 316.84 3168.4
σ = sqrt[(Σ(f(x) - mean)^2) / n] = sqrt[ (1987.44 + 1742.4 + 2353.4 + 392.04 + 642.88 +
2312.92 + 2293.76 + 3168.4) / 336] ≈ 8.05
To find the median, we need to find the interval that contains the 168th value (the halfway
point between the 336 frequencies). The cumulative frequency column tells us that the 168th
value falls within the 320-325 interval, which has a cumulative frequency of 132. The
interval width is 325-320 = 5, and we need to find how much of this interval contains the
168th value. To do so, we calculate:
p = (168 - 132) / 82 = 0.439
Median = lower limit of the interval + (p * interval width) = 320 + (0.439 * 5) = 322.195
quartile = lower limit of the interval + (p * interval width)
where p is the fractional part of (n * quartile number) / 4 and n is the total frequency.
For the first quartile (Q1), we need to find the interval that contains the 84th value (which is
(336 * 1) / 4). The cumulative frequency column tells us that the 84th value falls within the
310-315 interval, which has a cumulative frequency of 16 + 35 = 51. The interval width is
315-310 = 5, and we need to find how much of this interval contains the 84th value. To do so,
we calculate:
p = (84 - 51) / 81 = 0.407
Q1 = lower limit of the interval + (p * interval width) = 310 + (0.407 * 5) = 312.035
For the third quartile (Q3), we need to find the interval that contains the 252nd value (which
is (336 * 3) / 4). The cumulative frequency column tells us that the 252nd value falls within
the 325-330 interval, which has a cumulative frequency of 132 + 82 + 38 = 252. The interval
width is 330-325 = 5, and we need to find how much of this interval contains the 252nd
value. To do so, we calculate:
p = (252 - 132 - 82) / 38 = 0.842
Q3 = lower limit of the interval + (p * interval width) = 325 + (0.842 * 5) = 329.21
Therefore, the first quartile (Q1) is approximately 312.035 and the third quartile (Q3) is
approximately 329.21.
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.
The median for this grouped data is approximately 321.875, and the mean is approximately
322.195.
Since the mean and the median are relatively close in value, this suggests that the data is
fairly symmetrically distributed. This is also evident from the frequency polygon and
histogram, where we see that the distribution is somewhat bell-shaped, with the highest
frequencies occurring in the middle of the data range and decreasing as we move towards the
extremes.
However, there is a slight right skew in the distribution, as we can see from the frequency
polygon and histogram where the right tail extends further than the left tail. This skewness is
also reflected in the fact that the mean is slightly larger than the median, indicating that the
right tail of the distribution is pulling the mean towards it.
Overall, we can conclude that the distribution is roughly symmetric but slightly skewed to the
right.
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.Carry out an appropriate statistical test to determine whether the data is normally
distributed.
To test for normality, we can use the Shapiro-Wilk test, which is a commonly used statistical
test for normality.
However, since our data is grouped and we only have the frequencies for each interval, we
cannot...