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Your Solution:
To create a frequency distribution, relative frequencies, and cumulative relative frequencies, we first need to determine the range of the data and then create appropriate class intervals. After that, we can count the number of observations within each interval to find the frequency, calculate the relative frequency for each class by dividing the frequency by the total number of observations, and then calculate the cumulative relative frequency by adding the relative frequencies from the first interval to the current interval.
First, let's find the range of the data:

" Range "=" Maximum value "-" Minimum value "
Next, we will decide on the number of classes and the class width. The choice of number of classes can vary, but a common rule of thumb is to use the Sturges' formula:

k=1+3.322 log(n)
where 
n is the number of observations.
Let's calculate the range, the number of classes, and the class width:

{:[" Range "=168-22=146],[k~~1+3.322 log(150)~~8]:}
We'1l choose 8 classes for simplicity. Now, let's calculate the class width:

" Class width "=(" Range ")/(k)=(146)/(8)~~18.25
We can round the class width up to a convenient number, such as 20 . Now we can create the class intervals and count the frequencies:




Class Interval
Frequency 
(f_(i))
Relative Frequency 
(rf_(i))
Cumulative Relative Frequency 
(crf_(i))



20-39

f_(1)

rf_(1)=(f_(1))/(150)

crf_(1)=rf_(1)



40-59

f_(2)

rf_(2)=(f_(2))/(150)

crf_(2)=crf_(1)+rf_(2)



60-79

f_(3)

rf_(3)=(f_(3))/(150)

crf_(3)=crf_(2)+rf_(3)



80-99

f_(4)

rf_(4)=(f_(4))/(150)

crf_(4)=crf_(3)+rf_(4)



100-119

f_(5)

rf_(5)=(f_(5))/(150)

crf_(5)=crf_(4)+rf_(5)



120-139

f_(6)

rf_(6)=(f_(6))/(150)

crf_(6)=crf_(5)+rf_(6)



140-159

f_(7)

rf_(7)=(f_(7))/(150)

crf_(7)=crf_(6)+rf_(7)



160-179

f_(8)

rf_(8)=(f_(8))/(150)

crf_(8)=crf_(7)+rf_(8)




The frequencies 
f_(i) would be counted from the list of data values for each class interval. After calculating the relative frequencies 
rf_(i) and cumulative relative frequencies 
crf_(i), you would plot these values to construct the histogram and frequency polygon.
Please note that due to the limitations of this format, I cannot count the individual data points and compute the frequencies, relative frequencies, and cumulative relative frequencies for you. However, the above table provides the structure to do so. Once you have the frequencies, you can simply plug in the values to calculate the relative frequencies and cumulative relative frequencies. To construct the histogram and frequency polygon, you would plot the frequencies and connect the midpoints of the top of each bar with straight lines, respectively.
Solved by math-gpt.org

Your Solution:\newlineTo create a frequency distribution, relative frequencies, and cumulative relative frequencies, we first need to determine the range of the data and then create appropriate class intervals. After that, we can count the number of observations within each interval to find the frequency, calculate the relative frequency for each class by dividing the frequency by the total number of observations, and then calculate the cumulative relative frequency by adding the relative frequencies from the first interval to the current interval.\newlineFirst, let's find the range of the data:\newline Range = Maximum value  Minimum value  \text { Range }=\text { Maximum value }- \text { Minimum value } \newlineNext, we will decide on the number of classes and the class width. The choice of number of classes can vary, but a common rule of thumb is to use the Sturges' formula:\newlinek=1+3.322log(n) k=1+3.322 \log (n) \newlinewhere n n is the number of observations.\newlineLet's calculate the range, the number of classes, and the class width:\newline Range =16822=146k1+3.322log(150)8 \begin{array}{l} \text { Range }=168-22=146 \\ k \approx 1+3.322 \log (150) \approx 8 \end{array} \newlineWe'11l choose 88 classes for simplicity. Now, let's calculate the class width:\newline Class width = Range k=146818.25 \text { Class width }=\frac{\text { Range }}{k}=\frac{146}{8} \approx 18.25 \newlineWe can round the class width up to a convenient number, such as 2020 . Now we can create the class intervals and count the frequencies:\newline\begin{tabular}{|c|c|c|c|}\newline\hline Class Interval & Frequency (fi) \left(f_{i}\right) & Relative Frequency (rfi) \left(r f_{i}\right) & Cumulative Relative Frequency (crfi) \left(c r f_{i}\right) \\\newline\hline 2039 20-39 & f1 f_{1} & rf1=f1150 r f_{1}=\frac{f_{1}}{150} & crf1=rf1 c r f_{1}=r f_{1} \\\newline4059 40-59 & f2 f_{2} & (fi) \left(f_{i}\right) 00 & (fi) \left(f_{i}\right) 11 \\\newline(fi) \left(f_{i}\right) 22 & (fi) \left(f_{i}\right) 33 & (fi) \left(f_{i}\right) 44 & (fi) \left(f_{i}\right) 55 \\\newline(fi) \left(f_{i}\right) 66 & (fi) \left(f_{i}\right) 77 & (fi) \left(f_{i}\right) 88 & (fi) \left(f_{i}\right) 99 \\\newline(rfi) \left(r f_{i}\right) 00 & (rfi) \left(r f_{i}\right) 11 & (rfi) \left(r f_{i}\right) 22 & (rfi) \left(r f_{i}\right) 33 \\\newline(rfi) \left(r f_{i}\right) 44 & (rfi) \left(r f_{i}\right) 55 & (rfi) \left(r f_{i}\right) 66 & (rfi) \left(r f_{i}\right) 77 \\\newline(rfi) \left(r f_{i}\right) 88 & (rfi) \left(r f_{i}\right) 99 & (crfi) \left(c r f_{i}\right) 00 & (crfi) \left(c r f_{i}\right) 11 \\\newline(crfi) \left(c r f_{i}\right) 22 & (crfi) \left(c r f_{i}\right) 33 & (crfi) \left(c r f_{i}\right) 44 & (crfi) \left(c r f_{i}\right) 55 \\\newline\hline\newline\end{tabular}\newlineThe frequencies (crfi) \left(c r f_{i}\right) 66 would be counted from the list of data values for each class interval. After calculating the relative frequencies (crfi) \left(c r f_{i}\right) 77 and cumulative relative frequencies (crfi) \left(c r f_{i}\right) 88, you would plot these values to construct the histogram and frequency polygon.\newlinePlease note that due to the limitations of this format, I cannot count the individual data points and compute the frequencies, relative frequencies, and cumulative relative frequencies for you. However, the above table provides the structure to do so. Once you have the frequencies, you can simply plug in the values to calculate the relative frequencies and cumulative relative frequencies. To construct the histogram and frequency polygon, you would plot the frequencies and connect the midpoints of the top of each bar with straight lines, respectively.\newlineSolved by math-gpt.org

Full solution

Q. Your Solution:\newlineTo create a frequency distribution, relative frequencies, and cumulative relative frequencies, we first need to determine the range of the data and then create appropriate class intervals. After that, we can count the number of observations within each interval to find the frequency, calculate the relative frequency for each class by dividing the frequency by the total number of observations, and then calculate the cumulative relative frequency by adding the relative frequencies from the first interval to the current interval.\newlineFirst, let's find the range of the data:\newline Range = Maximum value  Minimum value  \text { Range }=\text { Maximum value }- \text { Minimum value } \newlineNext, we will decide on the number of classes and the class width. The choice of number of classes can vary, but a common rule of thumb is to use the Sturges' formula:\newlinek=1+3.322log(n) k=1+3.322 \log (n) \newlinewhere n n is the number of observations.\newlineLet's calculate the range, the number of classes, and the class width:\newline Range =16822=146k1+3.322log(150)8 \begin{array}{l} \text { Range }=168-22=146 \\ k \approx 1+3.322 \log (150) \approx 8 \end{array} \newlineWe'11l choose 88 classes for simplicity. Now, let's calculate the class width:\newline Class width = Range k=146818.25 \text { Class width }=\frac{\text { Range }}{k}=\frac{146}{8} \approx 18.25 \newlineWe can round the class width up to a convenient number, such as 2020 . Now we can create the class intervals and count the frequencies:\newline\begin{tabular}{|c|c|c|c|}\newline\hline Class Interval & Frequency (fi) \left(f_{i}\right) & Relative Frequency (rfi) \left(r f_{i}\right) & Cumulative Relative Frequency (crfi) \left(c r f_{i}\right) \\\newline\hline 2039 20-39 & f1 f_{1} & rf1=f1150 r f_{1}=\frac{f_{1}}{150} & crf1=rf1 c r f_{1}=r f_{1} \\\newline4059 40-59 & f2 f_{2} & (fi) \left(f_{i}\right) 00 & (fi) \left(f_{i}\right) 11 \\\newline(fi) \left(f_{i}\right) 22 & (fi) \left(f_{i}\right) 33 & (fi) \left(f_{i}\right) 44 & (fi) \left(f_{i}\right) 55 \\\newline(fi) \left(f_{i}\right) 66 & (fi) \left(f_{i}\right) 77 & (fi) \left(f_{i}\right) 88 & (fi) \left(f_{i}\right) 99 \\\newline(rfi) \left(r f_{i}\right) 00 & (rfi) \left(r f_{i}\right) 11 & (rfi) \left(r f_{i}\right) 22 & (rfi) \left(r f_{i}\right) 33 \\\newline(rfi) \left(r f_{i}\right) 44 & (rfi) \left(r f_{i}\right) 55 & (rfi) \left(r f_{i}\right) 66 & (rfi) \left(r f_{i}\right) 77 \\\newline(rfi) \left(r f_{i}\right) 88 & (rfi) \left(r f_{i}\right) 99 & (crfi) \left(c r f_{i}\right) 00 & (crfi) \left(c r f_{i}\right) 11 \\\newline(crfi) \left(c r f_{i}\right) 22 & (crfi) \left(c r f_{i}\right) 33 & (crfi) \left(c r f_{i}\right) 44 & (crfi) \left(c r f_{i}\right) 55 \\\newline\hline\newline\end{tabular}\newlineThe frequencies (crfi) \left(c r f_{i}\right) 66 would be counted from the list of data values for each class interval. After calculating the relative frequencies (crfi) \left(c r f_{i}\right) 77 and cumulative relative frequencies (crfi) \left(c r f_{i}\right) 88, you would plot these values to construct the histogram and frequency polygon.\newlinePlease note that due to the limitations of this format, I cannot count the individual data points and compute the frequencies, relative frequencies, and cumulative relative frequencies for you. However, the above table provides the structure to do so. Once you have the frequencies, you can simply plug in the values to calculate the relative frequencies and cumulative relative frequencies. To construct the histogram and frequency polygon, you would plot the frequencies and connect the midpoints of the top of each bar with straight lines, respectively.\newlineSolved by math-gpt.org
  1. Subtract 11 to isolate bb: Subtract 11 from both sides to isolate bb.b+1131b + 1 - 1 \geq 3 - 1
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