Question

Test scores X in a university Economics course that has 100 students every semester are known to have an mean value of 67 84 Further, it is known that approximately 95% of all class averages & are between 64 1 and 7151. From this information, approximate the standard deviation of the test scores X Select one a 187 00 ob 0.19 c. 1870.00 Od 1.87 e. 18.70

Answer 1

Given,

$\small \mu = 67.84$

$\small n = 100$

The z score corresponding to a 95% confidence interval is,

z = 1.96

We know that,

$\small Confidence\; Interval = \mu \pm z * \frac{\sigma} {\sqrt{n}}$

For the lower bound,

$\small 64.1 = 67.84- 1.96 * \frac{\sigma} {\sqrt{100}}$

$\small \sigma = 19.081$

For the upper bound,

$\small 71.5 = 67.84+1.96 * \frac{\sigma} {\sqrt{100}}$

$\small \sigma = 18.6734$

Therefore the approximate standard deviation of the test score X is,

e. 18.70