You are obsessed with how many views your pictures get. On average, you get 3022 views per post and always post exactly one picture per day. You’d like to have a month with 100000 or more total views. Find the approximate probability of this occurring in January if we assume view counts from day-to-day are independent.

The approximate probability of getting 100000 views or more in January if we assume view counts from day-to-day are independent = 0.22254

Step-by-step explanation:

January has 31 days.

The average number of views per day = 3022 views per day.

In terms of hourly basis, the average number of views = 3022/24 ≈ 126 views per hour

Then we need to find the probability that the number of views in January is equal to or exceeds 100000.

100000 views in January = 100000/31 = 3225.81 ≈ 3226 views per day

On an hourly basis, 3226 views per day ≈ 135 views per hour.

So, mean = λ = 126 views per hour

x = 135 views per hour.

Using Poisson's distribution function

P(X = x) = (e^-λ)(λˣ)/x!

P(X ≥ x) = Σ (e^-λ)(λˣ)/x! (Summation From x to the end of the distribution)

But it's easier to obtain

P(X < x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)

P(X ≥ x) = 1 - P (X < x)

Putting λ = 126 views/hour and x = 135 views/hour in the Poisson distribution formula calculator

P(X ≥ 135) = 0.22254

where is the triangle?

every 5 customers that walk in, only 2 buy popcorn.

step-by-step explanation:

basically 2/5.

brainliest would be appreciated.