Question

Assume a Normal distribution with an average return of​ 7% and a standard deviation of​ 2%. What is the probability of an actual return of​ (a) more than​ 11%; and​ (b) less than​ 5%?

Answer 1

Answer:

0.023 is the probability of an actual return of​ more than​ 11%.

0.159 is the probability of an actual return of​ less than​ 5%.                                              

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7%

Standard Deviation, σ = 2%

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

Formula:

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

a) P(more than​ 11%)

P(x > 11)

$P( x > 11) = P( z > \displaystyle\frac{11 - 7}{2}) = P(z > 2)$

$= 1 - P(z \leq 2)$

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

$P(x > 11) = 1 - 0.977 = 0.023 = 2.3\%$

Thus, 0.023 is the probability of an actual return of​ more than​ 11%.

b) P(less than​ 5%)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 7}{2}) = P(z < -1)$

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

$P(x < 5) =0.159 = 15.9\%$

Thus, 0.159 is the probability of an actual return of​ less than​ 5%.