Question

In the 1992 presidential election, Alaska's 40 election districts averaged 1982 votes per district for President Clinton. The standard deviation was 578. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.

a. What is the distribution of X? X ~ N(,)

b. Is 1982 a population mean or a sample mean? Select an answer Population Mean Sample Mean

c. Find the probability that a randomly selected district had fewer than 1978 votes for President Clinton.

d. Find the probability that a randomly selected district had between 2107 and 2224 votes for President Clinton.

e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.

Answer 1

a) The distribution of X is normal distribution i.e.  X ~ N( = 1982, = 578² )

b) 1982 is the population mean since it is the average value of the whole population.

c) P( X < 1978 ) =
= P( Z < -0.00692 )
=  0.4972393 ( from the Z tables )

d) P( 2107 < X < 2224) =
= P( 0.2163 < Z < 0.4187 )
= P( Z < 0.4187 ) - P( Z < 0.2163 )
= 0.6622823 - 0.585623
= 0.07666

e) Since the first quartile is attained at 25% of the data,
P( X < x ) = 0.25
= 0.25

= 0.25
From the Z-tables, when z = -0.67 the P( Z < z) = 0.25
Thus,    = -0.67
x = 1594.74

Hope this answers your query!