Neela has 11.5 yards of fabric. She will use 20% of the fabric to make a flag. How many yards of fabric will she use?

The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in the manufacturing sector.

elena-14-01-66 [18.8K]

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 $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for Switzerland = $30.67

$\sigma$ = 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 < $\bar X$ < $31.00)

P($30.00 < $\bar X$ < $31.00) = P( $\bar X$ < $31.00) - P( $\bar X$ $\leq$ $30.00)

P( $\bar X$ < $31) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{31-30.67}{\frac{4}{\sqrt{40} } }$ ) = P(Z < 0.52) = 0.69847

P( $\bar X$ $\leq$ $30) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{30-30.67}{\frac{4}{\sqrt{40} } }$ ) = P(Z $\leq$ -1.06) = 1 - P(Z < 1.06)

= 1 - 0.85543 = 0.14457

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.

Therefore, P($30.00 < $\bar X$ < $31.00) = 0.69847 - 0.14457 = 0.5539

(b) Suppose 32 manufacturing workers are selected randomly from across Japan.

Let $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for Japan = $20.20

$\sigma$ = 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( $\bar X$ > $21.00)

P( $\bar X$ > $21) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{21-20.20}{\frac{4}{\sqrt{32} } }$ ) = P(Z > 1.13) = 1 - P(Z < 1.13)

= 1 - 0.87076 = 0.12924

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.

(c) Suppose 47 manufacturing workers are selected randomly from across United States.

Let $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for United States = $23.82

$\sigma$ = 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( $\bar X$ < $22.80)

P( $\bar X$ < $22.80) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{22.80-23.82}{\frac{4}{\sqrt{47} } }$ ) = P(Z < -1.75) = 1 - P(Z $\leq$ 1.75)