Answers to fill in this table?????

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)

= 1 - 0.95994 = 0.04006

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.

3 0

3 years ago

What is the value of the sum?

Anna35 [415]

Answer:

-1/16

Step-by-step explanation:

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Devon purchased a new car valued at $16,000 that depreciated continuously at a rate of 35%. Its current value is $2,000. The equ

PilotLPTM [1.2K]

Cost =16000
rate of depreciation = 35%
written down value = 2000
depreciation for one year =16000*35% = 5600
5600*2 = 11200+5600*6\12 = 2000
the car is 2.5 years old. It has been depreciated for 2 whole years and half years.

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4 years ago

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A ball is drawn at random from a box containing 12 red,18

Lubov Fominskaja [6]

Answer:

For a: The probability of getting a red or blue ball is 0.48

For b: The probability of getting a white, blue or orange ball is 0.81

For c: The probability of getting neither white or orange ball is 0.48

Step-by-step explanation:

Probability is defined as the extent to which an event is likely to occurs. It is measured by the ratio of the favorable outcomes to the total number of possible outcomes.

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of favorable outcomes}}$ .......(1)

We are given:

Number of red balls in a box = 12

Number of white balls in a box = 18

Number of blue balls in a box = 19

Number of orange balls in a box = 15

Total balls in a box = [12 + 18 + 19 + 15] = 64

For a:

Number of favorable outcomes (ball must be red or blue) = [12 + 19] = 31

Total number of outcomes = 64

Putting values in equation 1, we get:

$\text{Probability of getting a red or blue ball}=\frac{31}{64}=0.48$

For b:

Number of favorable outcomes (ball must be white or blue or orange) = [18 + 19 + 15] = 52

Total number of outcomes = 64

Putting values in equation 1, we get:

$\text{Probability of getting a white or blue or orange ball}=\frac{52}{64}=0.81$

For c:

Number of favorable outcomes (ball must be white or orange) = [18 + 15] = 33

Total number of outcomes = 64

Putting values in equation 1, we get:

$\text{Probability of getting a white or orange ball}=\frac{33}{64}=0.52$

Probability of getting a ball which is neither white or orange = [1 - (Probability of getting a white or orange ball)] = [1 - 0.52] = 0.48

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3 years ago

The lines graphed below are perpendicular. The slope of the red line is -2.

ASHA 777 [7]

Answer:

1/2

Step-by-step explanation:

opposite reciprocal

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