Use the magnitudes, rounded to two decimal places, of the 100100 earthquakes included in the accompanying data set to construct a frequency distribution. Use a class width of 0.500.50 and begin with a lower class limit of 0.000.00. Does the frequency distribution appear to be a normal distribution? LOADING... Click the icon to view the earthquake magnitudes. Construct the frequency distribution. Magnitude Frequency 0.000.00-0.490.49 1010 0.500.50-0.990.99 3333 11-1.491.49 3535 1.501.50-1.991.99 1414 22-2.492.49 66 2.502.50-3.993.99 22 (Type integers or decimals. Do not round.) Does the frequency distribution appear to be a normal distribution? The frequency distribution could reasonably be a normal distribution because the frequencies start low, increase, and then decrease, and are roughly symmetric.

Step-by-step explanation:

Hello!

To arrange the data in a frequency table you have to determine the class intervals first. Normally you'd choose the number of intervals you want to work with and then calculate the class width by dividing the number ob observations, n, by the number of intervals, then, starting from the minimum value as the lower limit of the first interval, you add the width to calculate the upper interval. For the next interval, you use the upper limit from the previous one and add the width to determine the upper limit, and so on until you reach the desired number of intervals.

For this exercise the width was determined 0.50 and the lower limit for the first interval 0.00:

1st interval: 0.00+0.50= 0.50: [0.00; 0.50)

Then:

[0.50; 1.00)

[1.00; 1.50)

[1.50; 2.00)

[2.00; 2.50)

[2.50; 3.00)

Next is to arrange the data from least to greatest and count how many fit in each interval:

0 , 0,09, 0,18, 0,24, 0,31, 0,36, 0,39, 0,39, 0,41, 0,45, 0,55, 0,55, 0,63, 0,64, 0,64, 0,64, 0,64, 0,64, 0,69, 0,69, 0,71, 0,71, 0,73, 0,74, 0,78 ,0,79, 0,79, 0,79 ,0,83, 0,83, 0,83 , 0,85, 0,89, 0,91, 0,92, 0,92, 0,93,0,94, 0,98, 0,98 ,0,99, 0,99, 0,99,0,99, 1,01, 1,01, 1,01, 1,01, 1,22, 1,23,1,24, 1,24 , 1,24, 1,25 ,1,,25 , ,1,25,1,26 , 1,26, 1,27, 1,28 ,1,31 , 1,33 , 1,34 , 1,34 ,, 1,35, 1,35 , 1,38 , 1,39 , 1,41 , 1,42 , 1,42 , 1,43 , 1,44 , 1,44 , 1,45 , 1,46 , 1,49 , 1,49 , 1,56 , 1,56 , 1,62 , 1,63 , 1,65 , 1,65 , 1,75 , 1,77 , 1,78 , 1,78 , 1,82 , 1,85 , 1,96 , 1,98 , 2,19 , 2,21 , 2,23 , 2,24 , 2,47 , 2,49 , 2,95 , 2,97

See frequency table in attachment alongside with original data.

To check the distribution of the data I've made a QQ-plot with the Normal Quartiles (2nd attachment)

As you can see the Magnitude data is well-adjusted to the theoretical line for the normal distribution, this means that it is most likely that the data has a normal distribution.

I hope this helps!

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step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

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