Answer to Question #307999 in Statistics and Probability for Kelly

Question

In a national achievement test, the mean was found to be 62 and the standard deviation was 10. The scores also approximate the normal distribution.

1. What is the minimum score that belongs to the upper 10% of the group?
2. What are the two extreme scores outside of which 5% of the group are expected to fall?
3. What is the score that divides the distribution into two such that 75% of the cases is below it?
4. Estimate the range of scores that will include-

a. the middle 50% of the distribution.

b. the middle 95% of the distribution.

Answer 1

Mean $(\mu)$$= 62$

Standard deviation $(σ) = 10$

1. Minimum score for upper $10 \%$ of the group.

$=90\%$ and above.

$=P(0.9000)$

from distribution tables

Z for $P(0.9000) =1.28$

$Z=\dfrac{X-\mu}{σ}$ , $X=σZ+\mu$

$X=1.28*10+62$

$X=74.8\%$

2. Two extreme scores outside of which $5\%$ is expected to fall.

$=2.5\%$ and $97.5\%$

Z for P(0.9750) and P(0.0250)

Z = 1.96 and -1.96

$X=σZ+\mu$

$X(0.9750) = 1.96*10+62=81.6\%$

$X(0.0250) = -1.96*10+62=42.4\%$

3. Score that divides the group at the $75\%$

$Z for P(0.7500)= 0.68$

$X=σZ+\mu$

$X=0.68*10+62=68.8\%$

4.(a) scores that include the middle $50\%$

$=25\% and 75\%$

Z for $P(0.25) =-0.68$

$X=σZ+\mu$

$X=-0.68*10+62=55.2\%$

Z for $P(0.75)=0.68$

$X=σZ+\mu$

$X=0.68*10+62=68.8\%$

(b) score that will include the middle $95\%$

$=2.5\% and 97.5\%$

Z for $P(0.025) = -1.96$

$X=-1.96*10+62=42.4\%$

Z for $P(0.975) =1.96$

$X=1.96*10+62=81.6\%$