Describe the center and spread of the data using either the...

Describe the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice by constructing a box-and-whisker plot for the data. Be sure to include the scale. A. 47, 16, 70, 80, 28, 33, 91, 55, 60, 45, 86, 54, 30, 98, 34, 87, 44, 35, 64, 58, 27, 67, 72, 68, 31, 95, 37, 41, 97, 56, 49, 71, 84, 66, 45, 93 B. 40, 39, 37, 26, 25, 40, 35, 34, 26, 39, 42, 33, 26, 25, 34, 38, 41, 34, 37, 39, 32, 30, 22, 38, 36, 28, 27, 39, 34, 26, 36, 38, 25, 39, 23, 8

Transcript

00:01 So let us start with the concept which we are going to use here for this question.

00:06 Since we know the mean can be calculated as x bar is equals to the summation of observation that is xi over the total observation that is n.

00:16 So according to the question we are given the total observation that is n is equals to 36.

00:21 So first we need to find the mean range and standard deviation using below formula so that we can able to find the center and speed of the given data.

00:32 Okay so let us have a do this.

00:35 So by using this let us calculate first the mean x bar.

00:39 So this should be equals to.

00:40 So how did we calculate the mean that is sum of all the observation.

00:44 So summing all the observation will comes to 114 divided by total number of observation is 36.

00:50 So this will comes out equals to 58 .7 of 2.

00:54 Okay now the second term which we need to find is the standard deviation which is denoted by s and this is equals to under root over summation of xi minus of x bar the total square divided by the n minus of 1.

01:11 So let us start with this.

01:13 So this is equals to the under root over.

01:16 So first reading is 47 minus of the x bar we have just found 58 .7 to that whole square.

01:23 Okay this is what the first plus it will be up to that dash plus the last one that is 93 minus of 58 .7 of 2 that whole square divided by the n minus of 1 that means 36 minus of 1.

01:40 So after simplifying and solving this width comes as is equals to under root over 535 .0635 this would be equals to 23 .1314.

01:56 Now the third thing which we need to find is the range so that we can able to find the spread data and mean of the data.

02:06 Right so range is given by the maximum value minus of the minimum value.

02:14 So from the available data the maximum value is 98 minus minimum 16.

02:19 So range would comes out 82.

02:21 So therefore by the means of this for the given data the spread is nothing but equals to the range and that we have found to be as 82...

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