Question

Expand Previous Next Check 0/1 ptRetries 3Reattempts 99 In the 1992 presidential election, Alaska's 40 election districts averaged 2164 votes per district for President Clinton. The standard deviation was 580. (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 2164 a population mean or a sample mean? Select an answer c. Find the probability that a randomly selected district had fewer than 2091 votes for President Clinton. d. Find the probability that a randomly selected district had between 2217 and 2384 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to

the nearest whole number.

Answer 1

Here, sample size =40, sample mean = 2164 votes per district and The standard deviation was 580.

Since, X is normally distributed.

a. X~ N(2164, 580²) or X~ N(2164, 336400)

b. Since n = 40, which is small, it is a sample mean.

c. P(X<2091) = P(z < (2091-2164)/580) = P(z < -0.126) = 0.4483 (from normal table)

as

d. P(2217 <X< 2384) = P( (2217-2164)/580 < z < (2384-2164)/580) ) = P(0.09 < z < 0.379)= P(0.09 < z < 0.38)

= P( z < 0.38) - P(z<0.09) = 0.6480 - 0.5359 = 0.0121

e. Third quartile means 75% of the population covered,

or probability = 0.75 and corresponding z = 0.675

and, (x-2164)/580 = 0.67

so, x = 0.67*580 + 2164

x = 388.6+2164

x = 2552.6 is the third quartile

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