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

0.0014 is the required probability.

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* √(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(x-\mu)/(\sigma)`

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

`P( x > 1000000) \\\\= P( z > \displaystyle(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.