Problem 10 Easy Difficulty

The diameters of circles (in $\mathrm{mm}$ ) drawn in a design are given below:
$$
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Diameters } & 33-36 & 37-40 & 41-44 & 45-48 & 49-52 \\
\hline \text { No. of circles } & 15 & 17 & 21 & 22 & 25 \\
\hline
\end{array}
$$

Answer

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Intro Stats / AP Statistics

NCERT Class 11 - Math

Chapter 15

Statistics

Section 2

Measures of Dispersion

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Problem 10

Video Transcript

hell in this problem. We have given the diameter also circle and we are asked to calculate the standard deviation and mean diameter of a circle. So we have given the following data. So first of all we ever make this data continuous by subtracting 0.5 from the global limit And adding 0.5 to the a parliament. So we get the modified class that is 32.5- 36.5. 36.5- 40.5, 40.5- 44.5, 44.5 to 48.5 and 48.5- 52.5. Now we will find the mid values. That is exciting. It would be 34.5, 38.5. Well, the 2.5, 46.5 and 50.5 Let's assume the mean is equals two 42.5. Now we will find you I which is equals two x i minus assumed. Mean upon four therefore, is the height of the glass in the world. So we get you IS -2 minus one, zero one and two. Now we will find a freeway, it will be minus 30 minus 17, zero, 22 and 15. Similarly, we will find a fi you Y square be again 60 17, zero, 2200. The submission of FIEYS 25 and the submission of f i u i square is 199 As we know that men as equals two assumed men plus sigma, F I T Y opponent into it. By substituting the known values we get. Maine is equal to 42.5 plus 25 upon 100 into four, which is equals two 43.5. Now we will find standard deviation, which is equal to it's a bone in into under route and in two sigma, f i u y square minus sigma, f i u y whole square. By substituting the known values, we get four upon 102 Under route 102 99 -25 sq. It would be equals to run a .25 into Under route 19275, That is equals two one upon 25 into 138.83 It would be equals to 5.55. Therefore, we get Mean as equals two 43.5 and Standard Division As it was to 5.55.

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