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1,957.8 votes per district for a candidate. The standard deviation was 572.4. (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.

Part (a)
State the approximate distribution of X. (Enter your numerical values to one decimal place.)
X ~

,

Part (b)
Is 1,957.8 a population mean or a sample mean? How do you know?
A population mean, because all election districts are included.
A population mean, because only a sample of election districts are included.
A sample mean, because only a sample of election districts are included.
A sample mean, because all election districts are included.
Part (c)
Find the probability that a randomly selected district had fewer than 1,600 votes for this candidate. (Round your answer to four decimal places.)


Sketch the graph.
WebAssign Plot WebAssign Plot
WebAssign Plot WebAssign Plot

Write the probability statement.
P



Part (d)
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.)

Part (e)
Find the third quartile for votes for this candidate. (Round your answer up to the next vote.)
votes

Respuesta :

Part (a)
X
∼
N
(
1957.8
,
572.
4
2
)
X∼N(1957.8,572.4
2
)

Part (b)
1,957.8 is a sample mean because it is calculated from the 40 election districts in Alaska, which represent only a sample of all election districts.

Part (c)
To find the probability that a randomly selected district had fewer than 1,600 votes for this candidate, you need to calculate the z-score and then find the probability using the standard normal distribution table.

First, calculate the z-score:
z
=
1600
−
1957.8
572.4
=
−
0.6232
z=
572.4
1600−1957.8
​
=−0.6232

Now, find the probability using the standard normal distribution table:
P
(
X
<
1600
)
=
P
(
Z
<
−
0.6232
)
≈
0.2660
P(X<1600)=P(Z<−0.6232)≈0.2660

Part (d)
To find the probability that a randomly selected district had between 1,800 and 2,000 votes for this candidate, you need to calculate the z-scores for both values and then find the probability between those two z-scores.

First, calculate the z-scores:
For 1,800:
z
1
=
1800
−
1957.8
572.4
=
−
0.2766
z
1
​
=
572.4
1800−1957.8
​
=−0.2766
For 2,000:
z
2
=
2000
−
1957.8
572.4
=
0.0742
z
2
​
=
572.4
2000−1957.8
​
=0.0742

Now, find the probability:
P
(
1800
<
X
<
2000
)
=
P
(
−
0.2766
<
Z
<
0.0742
)
P(1800Using the standard normal distribution table, the probability is approximately 0.3094.

Part (e)
To find the third quartile for votes for this candidate, you need to find the z-score corresponding to the third quartile (which is 0.75) and then convert it back to the actual number of votes.

From the standard normal distribution table, the z-score corresponding to the third quartile is approximately 0.6745.

Now, use the z-score formula to find the vote count:
0.6745
=
X
−
1957.8
572.4
0.6745=
572.4
X−1957.8
​

Solving for
X
X, we get:
X
=
1957.8
+
0.6745
×
572.4
X=1957.8+0.6745×572.4
X
≈
2301.36
X≈2301.36

Rounding up to the next vote, the third quartile for votes for this candidate is approximately 2302 votes.

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