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3 years ago

Helppppp pleaseeee!!!!!!!! (Show the work)

Mathematics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

Step-by-step explanation:

ABD (half of the rectangle) is a right-angled triangle.

so, Pythagoras applies.

c² = a² + b²

c is the Hypotenuse (the baseline opposite of the 90 degree angle) and in our case the line BD.

a and b are the 2 sides enclosing the 90 degree angle. in our case here the lines AD and AB.

so,

BD² = AD² + AB² = (6-1)² + (5-2)² = 5² + 3² = 25 + 9 = 34

BD = sqrt(34)

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Answer:

(a) The probability that the sample average will be between $30.00 and $31.00 is 0.5539.

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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.

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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

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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)

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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)

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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)

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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.

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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)

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