Question

The SAT and ACT college entrance exams are taken by thousands of students each year. The mathematics portions of each of these exams produce scores that are approximately normally distributed. In recent years, SAT mathematics exam scores have averaged 480 with standard deviation 100. The average and standard deviation for ACT mathematics scores are 18 and 6, respectively. a An engineering school sets 550 as the minimum SAT math score for new students. What percentage of students will score below 550 in a typical year

Answer 1

Complete Question

The SAT and ACT college entrance exams are taken by thousands of students each year. The mathematics portions of each of these exams produce scores that are approximately normally distributed. In recent years, SAT mathematics exam scores have averaged 480 with standard deviation 100. The average and standard deviation for ACT mathematics scores are 18 and 6, respectively.

a An engineering school sets 550 as the minimum SAT math score for new students. What percentage of students will score below 550 in a typical year

b What would the engineering school set as a comparable standard on the ACT math test?

Answer:

a

$P(X < 550 ) = 75.8 \%$

b

The  minimum for ACT is  $x = 22.2$

Step-by-step explanation:

From the question we are told that

  The mean for  SAT is  $\mu_1 = 480$

   The standard deviation  for  SAT is  $\sigma_1 = 100$

   The mean for ACT is  $\mu_2 = 18$

    The standard deviation for $\sigma_2 = 6$

Generally the proportion  of  students will score below 550 in a typical year is mathematically represented as

    $P(X < 550 ) = P(\frac{X - \mu_1 }{\sigma_1} < \frac{550 - 480}{100} )$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

   $P(X < 550 ) = P(Z< 0.7)$

From the z table  the area under  0.7  to the left is

=>$P(Z< 0.7) = 0.75804$

So

  $P(X < 550 ) = 0.75804$

Converting to percentage

  $P(X < 550 ) = 0.75804 * 100$

  $P(X < 550 ) = 75.8 \%$

Generally the score which the  engineering school would set as a comparable standard on the ACT math test is mathematically evaluated as

     $P( X < x) = P(\frac{X - \mu_2 }{\sigma_2} < \frac{x - 18 }{6} ) = 0.75804$

=>  $P( X < x) = P(Z < \frac{x - 18 }{6} ) = 0.75804$

From the normal distribution table the critical value of  0.75804  is  

   $z = 0.7$

So

       $\frac{x- 18}{6} = 0.7$

=>    $x = 22.2$