Please assist me with 1.52!Thank you! 1.3 Describing Distributions with Numbers

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Please assist me with 1.52!Thank you!

1.3 Describing Distributions with Numbers 37 USE YOUR KNOWLEDGE 1.52 Find the IOR. Here are the scores on the first exam in an introduc- tory statistics course for 10 students: 83 74 93 85 75 97 93 55 92 81 STAT Find the in Find the interquartile range and use the 1.5 x 1QR rule to check for outliers. How low would the lowest score need to be for it to be an outlier according to this rule? Two variations on the basic boxplot can be very useful. The first, called a modified boxplot, uses the 1.5 × OR rule. The lines that extend out from the quartiles are terminated in whiskers that are 1.5 x IQR in length. Points beyond the whiskers are plotted individually and are classified as outliers according to the 1.5 x IQR rule. modified boxplot The other variation is to use two or more boxplots in the same graph to compare groups measured on the same variable. These are called side-by-side boxplots. The following example illustrates these two variations. side-by-side boxplots

Explanation / Answer

Q 1.52)

First we arrange the data in ascending order:

55,74,75,81,83,85,92,93,93,97

First Quartile : Q1 = (n+1)/4 = 11/4 =2.75 th value

First Quartile is between 2nd and 3rd value of the arranged data

Q1 = 2nd value+ 0.75(3rd value -2nd value) = 74+0.75(75-74)= 74.75

Third Quartile , Q3 = 3(n+1)/4 = 8.25th value

Thied Quartile is between 8th and 9th value of the arranged data

Q3 = 8th value+0.25(9th value - 8th value) = 93+0.25(93-93) = 93

Inter Quartile Range (IQR) = Q3- Q1 = 93-74.75 =18.25

The Inner and outer fence for outliers are

Inner Fence = Q1 -1.5IQR = 74.75-1.5*18.25 =47.375

Outer Fence = Q3 +1.5IQR = 93+1.5*18.25 = 120.375

All the data values are between inner and outer fence, hence there is no outlier

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