1) In a certain presidential election, Alaska\'s 40 election districts averaged
ID: 2922032 • Letter: 1
Question
1) In a certain presidential election, Alaska's 40 election districts averaged 1,955.8 votes per district for a candidate. The standard deviation was 572.5. (There are only 40 election districts in Alaska.) The distribution of the votes per district for one candidate was bell-shaped. Let X = number of votes for this candidate for an election district.
a) State the approximate distribution of X. (Enter your numerical values to one decimal place.)
b) Is 1,955.8 a population mean or a sample mean? How do you know?
c) Find the probability that a randomly selected district had fewer than 1,700 votes for this candidate. (Round your answer to four decimal places.)
d) Write the probability statement. P(____)
e) Find the probability that a randomly selected district had between 1,800 and 2,000 votes for this candidate. (Round your answer to four decimal places.)
f) Find the third quartile for votes for this candidate. (Round your answer up to the next vote.)
Explanation / Answer
a.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1955.8
standard Deviation ( sd )= 572.5
b.
it is population, since it cleary specify that there are only 40 election districts in Alaska. i.e we
have complete avalible data involved in the computation of mean
c.
P(X < 1700) = (1700-1955.8)/572.5
= -255.8/572.5= -0.4468
= P ( Z <-0.4468) From Standard Normal Table
= 0.3275
e.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1800) = (1800-1955.8)/572.5
= -155.8/572.5 = -0.2721
= P ( Z <-0.2721) From Standard Normal Table
= 0.3928
P(X < 2000) = (2000-1955.8)/572.5
= 44.2/572.5 = 0.0772
= P ( Z <0.0772) From Standard Normal Table
= 0.5308
P(1800 < X < 2000) = 0.5308-0.3928 = 0.138
f.
Q3 = P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.6745
P( x-u/s.d < x - 1955.8/572.5 ) = 0.75
That is, ( x - 1955.8/572.5 ) = 0.6745
--> x = 0.6745 * 572.5 + 1955.8 = 2341.9454
Q3 = 2341.9454 ~ 2342
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