Question

The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer thefollowing questions.a. What is the probability of completing the exam in one hour or less?b. What is the probability that a student will complete the exam in more than 60 minutesbut less than 75 minutes?c. Assume that the class has 60 students and that the examination period is 90 minutesin length. How many students do you expect will be unable to complete the exam inthe allotted time?

Answer 1

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(completing the exam in one hour or less)

P(x < 60)

$P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)$

Calculation the value from standard normal z table, we have,  

$P(x < 60) =0.023= 2.3\%$

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

$P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%$

c) P(completing the exam in more than 90 minutes)

P(x > 90)

$P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,  

$P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

$\dfrac{15.87}{100}\times 60 = 9.52\approx 10$

Approximately, 10 students of class will require more than 90 minutes to complete the test.