Question

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. What is the probability someone will spend no more than 30 minutes reading online national news reports

Answer 1

Answer:

0.118 is the probability someone will spend no more than 30 minutes reading online national news reports.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 49 minutes

Standard Deviation, σ = 16 minutes

We are given that the distribution of  time an individual reads is a bell shaped distribution that is a normal distribution.

Formula:

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

P(no more than 30 minutes)

$P( x \leq 610) = P( z \leq \displaystyle\frac{30 - 49}{16}) = P(z \leq -1.1875)$

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

$P(x \leq 30) = 0.118 = 11.8\%$

0.118 is the probability someone will spend no more than 30 minutes reading online national news reports.