Respuesta :
Answer:
a) Percentage of students scored below 300 is 1.79%.
b) Score puts someone in the 90th percentile is 638.
Step-by-step explanation:
Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.
(a) If the average test score is 510 with a standard deviation of 100 points.
To find : What percentage of students scored below 300 ?
Solution :
Mean [tex]\mu=510[/tex],
Standard deviation [tex]\sigma=100[/tex]
Sample mean [tex]x=300[/tex]
Percentage of students scored below 300 is given by,
[tex]P(Z\leq \frac{x-\mu}{\sigma})\times 100[/tex]
[tex]=P(Z\leq \frac{300-510}{100})\times 100[/tex]
[tex]=P(Z\leq \frac{-210}{100})\times 100[/tex]
[tex]=P(Z\leq-2.1)\times 100[/tex]
[tex]=0.0179\times 100[/tex]
[tex]=1.79\%[/tex]
Percentage of students scored below 300 is 1.79%.
(b) What score puts someone in the 90th percentile?
90th percentile is such that,
[tex]P(x\leq t)=0.90[/tex]
Now, [tex]P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90[/tex]
[tex]P(Z< \frac{t-\mu}{\sigma})=0.90[/tex]
[tex]\frac{t-\mu}{\sigma}=1.28[/tex]
[tex]\frac{t-510}{100}=1.28[/tex]
[tex]t-510=128[/tex]
[tex]t=128+510[/tex]
[tex]t=638[/tex]
Score puts someone in the 90th percentile is 638.
