The US Census Bureau reported the US population to be approximately 308,000,000 in 2010. It also reported that 6.5% of the popul
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Answer:
(a) The probability that the sample average will be between $30.00 and $31.00 is 0.5539.
(b) The probability that the sample average will exceed $21.00 is 0.12924.
(c) The probability that the sample average will be less than $22.80 is 0.04006.
Step-by-step explanation:
We are given that the hourly wage was $30.67 for Switzerland, $20.20 for Japan, and $23.82 for the U.S.
Assume that in all three countries, the standard deviation of hourly labor rates is $4.00.
(a) Suppose 40 manufacturing workers are selected randomly from across Switzerland.
Let = <u>sample average wage</u>
The z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean wage for Switzerland = $30.67
= standard deviation = $4.00
n = sample of workers selected from across Switzerland = 40
Now, the probability that the sample average will be between $30.00 and $31.00 is given by = P($30.00 < < $31.00)
P($30.00 < < $31.00) = P(
< $31.00) - P(
$30.00)
P( < $31) = P(
<
) = P(Z < 0.52) = 0.69847
P(
$30) = P(
) = P(Z
-1.06) = 1 - P(Z < 1.06)
= 1 - 0.85543 = 0.14457
<em>The above probability is calculated by looking at the value of x = 0.52 and x = 1.06 in the z table which has an area of 0.69847 and 0.85543 respectively.</em>
Therefore, P($30.00 < < $31.00) = 0.69847 - 0.14457 = <u>0.5539</u>
<u></u>
(b) Suppose 32 manufacturing workers are selected randomly from across Japan.
Let = <u>sample average wage</u>
The z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean wage for Japan = $20.20
= standard deviation = $4.00
n = sample of workers selected from across Japan = 32
Now, the probability that the sample average will exceed $21.00 is given by = P( > $21.00)
P( > $21) = P(
>
) = P(Z > 1.13) = 1 - P(Z < 1.13)
= 1 - 0.87076 = <u>0.12924</u>
<em />
<em>The above probability is calculated by looking at the value of x = 1.13 in the z table which has an area of 0.87076.</em>
<em />
(c) Suppose 47 manufacturing workers are selected randomly from across United States.
Let = <u>sample average wage</u>
The z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean wage for United States = $23.82
= standard deviation = $4.00
n = sample of workers selected from across United States = 47
Now, the probability that the sample average will be less than $22.80 is given by = P( < $22.80)
P( < $22.80) = P(
<
) = P(Z < -1.75) = 1 - P(Z
1.75)
= 1 - 0.95994 = <u>0.04006</u>
<em />
<em>The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.</em>