Question

Suppose the average length of time to finish a homework problem is 50 minutes. If a random sample of 9 students' times has standard deviation 6, what is the approximate probability the average time of the sampled students to complete the homework will be more than 53.7 minutes?

Answer 1

Solution :

$\sigma$
_T
= $\sigma$ / $\sqrt$ n = 6 / $\sqrt$ 9 = 2

P(
T
> 53.7) = 1 - P(
T
< 53.7)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
< (53.7 - 50) / 2]

= 1 - P(z < 1.85)

= 1 - 0.9678

= 0.0322

Probability = 0.0322