The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and

Question

3 years ago

15

that the times are normally distributed. What is the probability someone will spend no more than 30 minutes reading online national news reports

1 answer:

Olin [163] · Answer 1 · 2022-08-26T19:48:01+03:00

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.