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
[tex]\mu = 180000*5 = 900000\text{ per week}[/tex]
Standard Deviation, σ = 15000 per day
[tex]\sigma = 15000\times \sqrt{5} = 33541.02\text{ per week}[/tex]
We are given that the distribution of number of viewers is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(number of viewers is more than 1000000 in a week)
[tex]P( x > 1000000) \\\\= P( z > \displaystyle\frac{1000000 - 900000}{33541.02}) = P(z >2.9814)[/tex]
[tex]= 1 - P(z \leq 2.9814)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 610) = 1 - 0.9986 =0.0014[/tex]
0.0014 is the probability that the number of viewers is more than 1000000 in a week.
