Question

The SAT are scored so as to yield a mean of 500 and a standard deviation of 100. These scores are close to being normally distributed. An exclusive club wishes to invite those scoring in the top 5% on the College Boards to join. a. Find percentages of scores that is between 480 and 590. b. What score is required to be invited to join the club? c. What score separates the top 60% of the population from the bottom 40%? What do we call this value? (hint: the score that separates the top 20% of the population f

Answer 1

Complete Question

The SAT are scored so as to yield a mean of 500 and a standard deviation of 100. These scores are close to being normally distributed. An exclusive club wishes to invite those scoring in the top 5% on the College Boards to join. a. Find percentages of scores that is between 480 and 590. b. What score is required to be invited to join the club? c. What score separates the top 60% of the population from the bottom 40%? What do we call this value? (hint: the score that separates the top 20% of the population from the bottom 80% is called the 80th percentile)

Answer:

a

 $P(480 < X < 590) = 0.3952$  

b  

  $x = 664.5$

c

 $x_1 = 474.67$

The value is called the $40^{th}$ percentile

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 500$

The standard deviation is $\sigma = 100$

Generally the percentage of the score that are between 480 and 590 is mathematically represented as

$P(480 < X < 590) = P(\frac{ 480 - 500}{ 100} < \frac{ X - \mu }{ \sigma } < \frac{ 590 - 500}{ 100} )$

Here $\frac{ X -\mu}{\sigma } = Z$

=> $P(480 < X < 590) = P(\frac{ 480 - 500}{ 100} < Z < \frac{ 590 - 500}{ 100} )$

=> $P(480 < X < 590) = P(-0.2< Z < 0.9 )$

=> $P(480 < X < 590) = P(Z < 0.9 ) - P(Z$

From the z table (reference calculator dot net )

$P(Z < 0.9 )= 0.81594$

And

$P(Z < -0.2 )= 0.42074$

=> $P(480 < X < 590) = 0.81594 - 0.42074$

=> $P(480 < X < 590) = 0.3952$

Converting to percent

$P(480 < X < 590) = 0.3952 * 100$

$P(480 < X < 590) = 39.52\%$

The percentage invited to join the club is the top 5% , this can be mathematically represented as

$P(X > x ) = 0.05$

Here x is the score that is within the top 5%

=> $P(X > x) = P( \frac{X - \mu }{\sigma } > \frac{ x - 500}{100} )= 0.05$

Generally the critical value of 0.05 from the normal distribution table is

$Z_{0.05} = 1.645$

So

$\frac{ x - 500}{100} = 1.645$

=> $x - 500 = 164.5$

=> $x = 664.5$

Generally the probability of having a score that separates the top 60% of the population from the bottom 40%

$P(X > x_1 ) = 0.60$

=> $P(X > x_1) = P(\frac{ X - \mu }{\sigma } > \frac{x - 500}{100} ) = 0.60$

The critical value of 0.60 from the normal distribution table is

$Z_{0.60} = - 0. 2533$

So

$\frac{x_1 - 500}{100} = -0.2533$

=> $x_1 = 474.67$

From the hint given we can see that this value is called the $40^{th}$ percentile