Zoagliaj

2022-07-17

A class of 135 students took a final examination in mathematics. The scores were normally distributed with a mean score of 68% and a standard deviation of 8.5%. Determine the percentile rank of each of the following people:

(a) Jennifer, who scored 78 %

(b) Steven, who got 55 %

(c) Jessica, who was very happy with her mark of 89 %

(a) Jennifer, who scored 78 %

(b) Steven, who got 55 %

(c) Jessica, who was very happy with her mark of 89 %

Osvaldo Crosby

Beginner2022-07-18Added 12 answers

Given :

n = 135

$\text{Mean}=\mu =68$

$\text{Standard deviation}=\sigma =8.5$

a) $Z=\frac{X-\mu}{\sigma}=\frac{78-68}{8.5}=1.18$

P(Z < 1.18) = 0.8810 (from standard normal table)

Percentile rank = 88.10%

b) $Z=\frac{55-68}{8.5}=1.53\phantom{\rule{0ex}{0ex}}P(Z<-1.53)=0.0630\phantom{\rule{0ex}{0ex}}\Rightarrow \text{Percentile rank}=6.3\mathrm{\%}$

c) $Z=\frac{89-68}{8.5}=2.47\phantom{\rule{0ex}{0ex}}P(Z<2.47)=0.9932\phantom{\rule{0ex}{0ex}}\Rightarrow \text{Percentile rank}=99.32\mathrm{\%}$

n = 135

$\text{Mean}=\mu =68$

$\text{Standard deviation}=\sigma =8.5$

a) $Z=\frac{X-\mu}{\sigma}=\frac{78-68}{8.5}=1.18$

P(Z < 1.18) = 0.8810 (from standard normal table)

Percentile rank = 88.10%

b) $Z=\frac{55-68}{8.5}=1.53\phantom{\rule{0ex}{0ex}}P(Z<-1.53)=0.0630\phantom{\rule{0ex}{0ex}}\Rightarrow \text{Percentile rank}=6.3\mathrm{\%}$

c) $Z=\frac{89-68}{8.5}=2.47\phantom{\rule{0ex}{0ex}}P(Z<2.47)=0.9932\phantom{\rule{0ex}{0ex}}\Rightarrow \text{Percentile rank}=99.32\mathrm{\%}$

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