Answer:
a. X~N(1898;358801)
b. population mean
c. P(X<1904)= 0.50399
d. P(2006≤X≤2197)= 0.12004
e. Third quantile: x₀= 2299.33
Step-by-step explanation:
Hello!
The variable of interest is
X: number of votes for President Clinton for an election district.
Population: districts of Alaska
Mean μ=1898 votes
Standard deviation δ= 599 votes ⇒ Variance δ²= 358801
The distribution for the votes per district for president Clinton is bell-shaped.
a. The distribution for this variable is normal: X~N(1898;358801)
Mean μ=1898 votes
b. 1898 is the population mean.
Since the population of interest is the "40 districts of Alaska" and this value corresponds to the average number of votes for Clinton in all of the Alaskan districts, then it is the population mean.
c. Symbolically:
P(X<1904)
To find this probability you have to use the standard normal tables, so first is to standardize or "transform" the value of X into a value of Z using: Z=(X-μ)/δ ~N(0;1)
P(X<1904)= P(Z<(1904-1898)/599)= P(Z<0.01) = 0.50399
d. Symbolically:
P(2006≤X≤2197)= P(X≤2197)-P(X≤2006)
P(Z≤(2197-1898)/599)-P(X≤(2006-1898)/599)= P(Z≤0.50)-P(X≤0.18)= 0.69146-0.57142= 0.12004
e. The third quantile is the value of the variable that separates the bottom 75% of the distribution from the top 25%, symbolically:
P(X≤x₀)=0.75
To find this value you have to look in the body of the Z-table the probability of 0.75 and then it's corresponding Z-value (z₀), then using that Z-value you have to "reverse" the standardization and clear the value of x₀ from the formula of Z:
z₀= 0.67
z₀= (x₀-μ)/δ
δ*z₀= x₀-μ
x₀= (δ*z₀)+μ
x₀= (599*0.67)+1898
x₀= 2299.33
I hope this helps!
