The way the answer was worked out for the example problem \/ exercise here was as follows:<\/p>\n

 <\/p>\n

We need the numerical labels for the seven “notches” or “tick marks” on the horizontal axis.<\/p>\n

 <\/p>\n

The middle of these seven notches or tick marks is x = 64.3\u00a0 ( which is the mean here )<\/p>\n

 <\/p>\n

Then we do these calculations to find the other six numerical labels for the tick marks:<\/p>\n

 <\/p>\n

64.3\u00a0 –\u00a0 2.62\u00a0 –\u00a0 2.62\u00a0 –\u00a0 2.62\u00a0 =\u00a0 56.44<\/p>\n

 <\/p>\n

64.3\u00a0 –\u00a0 2.62\u00a0 –\u00a0 2.62\u00a0 =\u00a0 59.06<\/p>\n

 <\/p>\n

64.3\u00a0 –\u00a0 2.62\u00a0 =\u00a0 61.68<\/p>\n

 <\/p>\n

64.3\u00a0 +\u00a0 2.62\u00a0 =\u00a0 66.92<\/p>\n

 <\/p>\n

64.3\u00a0 +\u00a0 2.62\u00a0 +\u00a0 2.62\u00a0 =\u00a0 69.54<\/p>\n

 <\/p>\n

64.3\u00a0 +\u00a0 2.62\u00a0 +\u00a0 2.62\u00a0 +\u00a0 2.62\u00a0 =\u00a0 72.16<\/p>\n

 <\/p>\n

So that the labels ( and these are x values and NOT z values ) for the seven tick marks from left to right in the picture then are:<\/p>\n

 <\/p>\n

56.44\u00a0\u00a0\u00a0\u00a0\u00a0 59.06\u00a0\u00a0\u00a0\u00a0\u00a0 61.68\u00a0\u00a0\u00a0\u00a0\u00a0 64.3\u00a0\u00a0\u00a0\u00a0\u00a0 66.92\u00a0\u00a0\u00a0\u00a0\u00a0 69.54\u00a0\u00a0\u00a0\u00a0\u00a0 72.16<\/p>\n

 <\/p>\n

And so when answering the question having to do with the percent of women whose heights are between 59.06 inches and 64.3 inches, what we see from the picture in the last slide above is that 13.5% occurs between x = 59.06 and x = 61.68 and that another 34% occurs between x = 61.68 and x = 64.3, and so that therefore the total percentage occurring between x = 59.06 and x = 64.3 is 13.5% + 34%\u00a0 =\u00a0 47.5%<\/p>\n

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Thanks Friends and Be Well this Week 3 !!<\/p>\n

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\ud83d\ude42<\/p>\n

Hi Friends and Greetings !!<\/p>\n

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As you have been finding out in your Weeks 1-3 Knewton Homework assignments, when there is a suspected outlier in the sample data, generally we shift towards preferring to use the median as a good measure of so-called central tendency.<\/p>\n

 <\/p>\n

On the other hand, if we don’t have any suspected outliers in the sample data, then we very often stick with preferring to use the mean as a good measure of central tendency.<\/p>\n

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When we say that we start to talk about and learn about distributions using shape, center, and spread, then what central tendency is – it is the fancy phrase or expression for “center.”<\/p>\n

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And the fancy phrase or expression for “spread” is “dispersion.”<\/p>\n

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Thanks Friends and I hope that you can conclude the Weeks 1-3 Knewton Homework assignments as quickly as possible and then try to dive into the Week 4 Knewton Homework assignments as quickly as is practical too !<\/p>\n

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Good Luck and Best Wishes Friends and THANK YOU for your hard work and attention !!<\/p>\n

MATH 225N Week 3 Discussion: Measures of Central Tendency and Variation<\/span><\/p>\n