Question

In a large lecture course, the scores on the final examination followed the normal curve closely. The average score was 60 points and one fourth of the class scored between 50 and 70 points. The standard deviation of the scores was

Answer 1

Answer:

The value is  $\sigma = 31.35$

Step-by-step explanation:

From the question we are told that

    The  average score is  $\mu = 60$

Generally the probability that  $\frac{1}{4} = 0.25$  of the class score is between 50 and 70 points is mathematically represented as

    $P(50 < X < 70)= P(\frac{50 - 60 }{\sigma } < \frac{X - \mu }{\sigma} < \frac{70 - 60}{\sigma } ) = 0.25$

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

So

       $P(50 < X < 70)= P(\frac{-10 }{\sigma } < Z < \frac{10}{\sigma } ) = 0.25$

=>   $P(50 < X < 70)= P(Z < \frac{10 }{\sigma }) -(Z < \frac{-10}{\sigma } )= 0.25$

=>  $P(50 < X < 70)= 2P(Z < \frac{10 }{\sigma }) = 0.25$

From the normal distribution table the critical value of  0.25 for a two tailed test is z= 0.318

So

      $\frac{10}{\sigma } = 0.319$

=>  $\sigma = 31.35$