Question

The amount of time devoted to studying statistics each week by students who achieve a grade of A in the course is a normally distributed random variable with a mean of 7.5 hours and a standard deviation of 2.1 hours.
a. What proportion of A students study for more than 10 hours per week?
b. Find the probability that an A student spends between 7 and 9 hours studying.
c. What proportion of A students spend fewer than 3 hours studying?
d. What is the amount of time below which only 5% of all A students spend studying?

Answer 1

Answer:

a) 11.7% of students study for more than 10 hours per week.    

b) 35.6% of student spends between 7 and 9 hours studying.

c) 1.6% of  students spend fewer than 3 hours studying.

d) 5% of all A students spend studying 4.0455 or less hours during a week.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.5 hours

Standard Deviation, σ = 2.1 hours

We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(students study for more than 10 hours per week)

P(x > 10)

$P( x > 10) = P( z > \displaystyle\frac{10 - 7.5}{2.1}) = P(z > 1.1904)$

$= 1 - P(z \leq 1.1904)$

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

$P(x > 10) = 1 - 0.883 = 0.117 = 11.7\%$

b) P(student spends between 7 and 9 hours studying.)

$P(7 \leq x \leq 9) = P(\displaystyle\frac{7 - 7.5}{2.1} \leq z \leq \displaystyle\frac{9-7.5}{2.1}) = P(-0.2380 \leq z \leq 0.7142)\\\\= P(z \leq 0.7142) - P(z < -0.2380)\\= 0.762 - 0.406 = 0.356 = 35.6\%$

$P(7 \leq x \leq 9) = 35.6\%$

c)  P(students spend fewer than 3 hours studying)

$P(x < 3) = P(z < \displaystyle\frac{3-7.5}{2.1}) = P(z < -2.1428)$

Calculating the value from the standard normal table we have,

$P( x < 3) =0.016 = 1.6\%$

d)  P(X < x) = 0.05

We have to find the value of x such that the probability is 0.035

$P( X < x) = P( z < \displaystyle\frac{x - 7.5}{2.1})=0.05$  

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

$\displaystyle\frac{x - 7.5}{2.1} = -1.645\\x = 4.0455$  

5% of all A students spend studying 4.0455 or less hours during a week.