### Calculate upper quartile, standard score and standard deviation of a stem-and-leaf diagram

Question Sample Titled 'Calculate upper quartile, standard score and standard deviation of a stem-and-leaf diagram'

The stem-and-leaf diagram below shows the distribution of the scores (in marks) of a group of students in a test. Jack gets the highest score in the test.

Stem (tens)Leaf (units)
 ${4}$ ${5}$ ${6}$ ${7}$ ${8}$
 ${2}$ ${2}$ ${2}$ ${5}$ ${4}$ ${4}$ ${5}$ ${1}$ ${2}$ ${3}$ ${3}$ ${8}$ ${9}$ ${1}$ ${6}$ ${6}$ ${0}$ ${1}$ ${5}$ ${6}$

 Which of the following is/are true? I. The upper quartile of the distribution is ${54}$ marks. II. The standard score of Jack in the test is lower than ${1.4}$ . III. The standard deviation of the distribution is greater than ${12}$ marks.

A
III only
B
II only
C
II and III only
D
I and II only

 I.  ${42},{42},{42},{45},{54},{54},{55},{61},{62},{63},{63},{68},{69},{71},{76},{76},{80},{81},{85},{86}$ Separate the data into two halves,  Median ${\left({Q}_{{2}}\right)}=\dfrac{{{63}+{63}}}{{2}}={63}$ $\dfrac{{{10}^{{\text{th}}}\text{datum}+{11}^{{\text{th}}}\text{datum}}}{{2}}$ Further separate the upper half and lower half into two halves,  Lower quartile ${\left({Q}_{{1}}\right)}=\dfrac{{{54}+{54}}}{{2}}={54}$ $\dfrac{{{5}^{{\text{th}}}\text{datum}+{6}^{{\text{th}}}\text{datum}}}{{2}}$ Upper quartile ${\left({Q}_{{3}}\right)}=\dfrac{{{76}+{76}}}{{2}}={76}$ $\dfrac{{{15}^{{\text{th}}}\text{datum}+{16}^{{\text{th}}}\text{datum}}}{{2}}$ I is false.  II. Using base mode of the calculator, calculate the mean and stardard score first. Mean ${\left(\mu\right)}={63.75}$ Stardard deviation ${\left(\sigma\right)}={14.063694393721729}\approx{14.064}$ ∴   Standard score of Jack$=\dfrac{{{86}-{63.75}}}{{14.064}}\approx{1.582}$ stardard score$=\dfrac{{{x}-\mu}}{\sigma}$ II is false.  III. Stardard deviation ${\left(\sigma\right)}\approx{14.064}$ III is true.

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