Answer:
1096750 impressions
Step-by-step explanation:
Given that :
Mean = 850,000
Standard deviation = 150,000
If we assume that X should be the numbers of impressions created;
Then ;
[tex]X \approx N (\mu , \sigma^2)[/tex]
Now ; representing x as the value for the number of impression needed ; Then ;
[tex]P(X>x) = 0.95[/tex]
[tex]P(\dfrac{X- \mu}{\sigma} > \dfrac{x -850000}{150000}) = 0.95[/tex]
[tex]P(Z> \dfrac{ x -850000}{150000}) = 0.95[/tex]
From normal tables:
[tex]P(Z >1.645) = 0.95[/tex]
[tex]\dfrac{x - 850000}{150000} =1.645[/tex]
(x- 850000) = 1.645(150000)
x - 850000 = 246750
x = 246750 + 850000
x = 1096750 impressions