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Question 1 of 4
1. Question
If a set of scores have a mean of `11` and a standard deviation of `0.9`, sketch the normal curve then find the zscore for `12`.Hint
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`Z` score
`Z=(xbarx)/S`Given Values
Mean `barX= 11`Standard Deviation`=0.9`First, complete the labels of the bell curve by using the mean and standard deviation.For example, start with the mean, `11`. Then add and subtract `0.9` to get the values `1` standard deviation above and below the mean.Keep adding and subtracting the standard deviation until the labels are completed.Add the zscore values.Input the values to the zscore formula to solve for the zscore.`Z` `=` `(xbarx)/S` `=` `(1211)/0.9` Substitute values `=` `1.11` Simplify `1.11` 
Question 2 of 4
2. Question
Given that the mean is `78.4` and the standard deviation is `3.2`, find the zscores for `68.2` and `82.4`.
`68.2` zscore: (3.19, 3.1875)`82.4` zscore: (1.25)
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`Z` score
`Z=(xbarx)/S`Given Values
Mean `barX= 78.4`Standard Deviation`=3.2`First, complete the labels of the bell curve by using the mean and standard deviation.For example, start with the mean, `78.4`. Then add and subtract `3.2` to get the values `1` standard deviation above and below the mean.Keep adding and subtracting the standard deviation until the labels are completed.Add the zscore values.Input the values to the zscore formula to solve for the zscore of `68.2`.`Z` `=` `(xbarx)/S` `=` `(68.278.4)/3.2` Substitute values `=` `(10.2)/3.2` Simplify `=` `3.19` Do the same to solve for the zscore of `82.4`.`Z` `=` `(xbarx)/S` `=` `(82.478.4)/3.2` Substitute values `=` `4/3.2` Simplify `=` `1.25` `68.2` zscore: `3.19``82.4` zscore: `1.25` 

Question 3 of 4
3. Question
Jack had a test for both Math and English. Given the following information, in which of the two subjects did he perform better?Hint
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`Z` score
`Z=(xbarx)/S`First, find the zscore for Math.`Z` `=` `(xbarx)/S` `=` `(8662)/13` Substitute values `=` `(24)/13` `=` `1.85` First, find the zscore for English.`Z` `=` `(xbarx)/S` `=` `(8673)/6` Substitute values `=` `13/6` `=` `2.17` In this problem, the better score is determined by how further away it is from the right of the mean.
Visualizing the zscores, we have:Therefore, Jack performed better in EnglishEnglish 
Question 4 of 4
4. Question
Women’s heights in a survey had a mean of `166.8`cm, a standard deviation of `3.1`cm, and is normally distributed.(a) Where do most heights almost certainly lie?(b) Find the zscore for the height of `173.5`cm.(c) Find the height for the zscore of `1.8`.
(a) (157.5)cm to (176.1)cm(b) zscore: (2.16)(c) height: (161.22)cm
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`Z` score
`Z=(xbarx)/S`Given Values
Mean `barX= 166.8` `cm`Standard Deviation`=3.1` `cm`First, complete the labels of the bell curve by using the mean and standard deviation.For example, start with the mean, `166.8` `cm`. Then add and subtract `3.1` `cm` to get the values `1` standard deviation above and below the mean.Keep adding and subtracting the standard deviation until the labels are completed.Add the zscore values.Solution for (a)`99.7%` of scores lie within `3` standard deviations away from the mean.Hence, almost all heights measure `157.5` `cm` to `176.1` `cm`For part (b), input the values to the zscore formula to solve for the zscore of the height, `173.5` `cm`.`Z` `=` `(xbarx)/S` `=` `(173.5166.8)/3.1` Substitute values `=` `6.7/3.1` Simplify `=` `2.16` The actual height of `173.5`cm is equivalent to a zscore of `2.16`.For part (c), input the known values into the zscore formula to solve for the corresponding height of the zscore, `1.8`.`Z` `=` `(xbarx)/S` `1.8` `=` `(x166.8)/3.1` Substitute values `1.8``xx3.1` `=` `(x166.8)/3.1``xx3.1` Multiply both sides by `3.1` `5.58` `=` `x166.8` Simplify `5.58``+166.8` `=` `x166.8``+166.8` Add `166.8` to both sides `161.22` `=` `x` `x` `=` `161.22` `cm` The zscore of `1.8` is equivalent to a height of `161.22`cm.(a) `157.5`cm to `176.1`cm(b) zscore: `2.16`(c) height: `161.22`cm 