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An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blends of gasoline. Construct a frequency distribution and histogram for these data. Use 8 bins.

90.9, 98.6, 89.6, 92.6, 91.8, 92.0, 92.3, 90.9
94.3, 88.3, 91.6, 82.5, 89.0, 89.8, 85.6, 90.2
83.1, 89.4, 92.1, 88.9, 92.0, 91.3, 87.0, 88.9
89.9, 92.1, 90.4, 86.7, 96.8, 93.8, 88.1, 92.5
91.9, 88.8, 89.3, 89.7, 91.3, 97.0, 89.1, 90.5
90.1, 89.6, 87.6, 94.9, 92.8, 84.1, 91.9, 92.6
92.4, 87.7, 95.7, 88.5, 90.5, 90.1, 89.6, 91.1
91.2, 85.6, 91.7, 88.9, 88.6, 89.0, 89.0, 88.2
91.1, 90.6, 93.5, 89.1, 90.8, 91.4, 95.3, 88.6
99.3, 91.9, 91.5, 94.2, 91.4, 95.0, 91.0, 88.9
90.1, 89.7, 85.6

Answer :

Answer:

I will be attaching files to explain the procedure

Step-by-step explanation:

From the word document, we can see the data was first arranged, and then a frequency distribution table was created, and cummulative frequency calculated.

The class interval used was 2.5 and the cummulative frequency was 83 ( this is the total amount of observations in the data).

The histogram was improvised and I will attach a picture of it.

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Rewritten by : Barada

Final answer:

A frequency distribution can be created by sorting the data into 8 bins; these intervals and corresponding frequencies are then used to create a histogram.

Explanation:

First, to create a frequency distribution, we need to sort the data and categorize it into bins. Given that we're asked to use 8 bins, we'll first find the range, which is the maximum value minus minimum value (99.3-82.5 = 16.8). Then, we would divide this range by the bin number (16.8/8 = 2.1). We can round it to 2 for simplicity. So, our bins would be: 82.5-84.5, 84.6-86.6, 86.7-88.7, 88.8-90.8, 90.9-92.9, 93.0-95.0, 95.1-97.1, 97.2-99.3. To create a histogram, one would plot these bins (intervals) on the x-axis and the frequencies on the y-axis. Each bin would be represented as a bar, the height of which corresponds to its frequency.

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