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As we already have the model that describes the change of the population in Italy in terms of the years that have elapsed, we only have to replace the conditions that are requested in that equation.
Therefore to find the population of Italy in the year 2000 (t = 10 years) substitute t = 10 in the equation and have:
million people
To find the population of Italy in 2008 (t = 18 years)
substitute t = 18 in the equation and have:
million people
To predict the population in Italy for 2015 and 2020 with this model, we substitute in the equation t = 25 and t = 30
t = 25
million people
t = 30
million people
We have the mean, μ = 20 and the standard deviation, σ = 4
X ≈ N(20, 4²)
We want P(X<15)
Standardising gives
P(X<15) = P(Z<
)
P(X<15) = P(Z< -1.25) ⇒ When the value of the z-score is negative, we will read the value of z<1.25 on the z-table then subtract the answer from 1
P(X<15) = 1 - P(Z<1.25)
P(X<15) = 1 - 0.8944
P(X<15) = 0.1056
Hence, the probability that pizza takes less than 15 minutes to be delivered is 0.1056 = 10.56%
X ≈ N(20, 4²)
We want P(X<15)
Standardising gives
P(X<15) = P(Z<
P(X<15) = P(Z< -1.25) ⇒ When the value of the z-score is negative, we will read the value of z<1.25 on the z-table then subtract the answer from 1
P(X<15) = 1 - P(Z<1.25)
P(X<15) = 1 - 0.8944
P(X<15) = 0.1056
Hence, the probability that pizza takes less than 15 minutes to be delivered is 0.1056 = 10.56%
therefore
Other method:
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