Question

g a local tv network estimates that the avergae number of viewers during primetime on weekdays is normalling distributed with a mean of 180000 viewers per day and a standard deviation of 15000 f we pick a week, what is the probability that the numver of viewers is mroe than 1000000

Answer 1

Answer:

0.0014 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 180000 per day

$\mu = 180000*5 = 900000\text{ per week}$

Standard Deviation, σ = 15000 per day

$\sigma = 15000\times \sqrt{5} = 33541.02\text{ per week}$

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

Formula:

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

P(number of viewers is more than 1000000 in a week)

$P( x > 1000000) \\\\= P( z > \displaystyle\frac{1000000 - 900000}{33541.02}) = P(z >2.9814)$

$= 1 - P(z \leq 2.9814)$

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

$P(x > 610) = 1 - 0.9986 =0.0014$

0.0014 is the probability that the number of viewers is more than 1000000 in a week.