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Alex73 [517]
3 years ago
15

According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and ente

rtainment is $2,060. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $495. (Round your z-score computation to 2 decimal places and final answers to 2 decimal places.)
What percent of the adults spend more than $2,575 per year on reading and entertainment?

What percent spend between $2,575 and $3,300 per year on reading and entertainment?

What percent spend less than $1,225 per year on reading and entertainment?
Mathematics
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

A) P(x> 2575) = 14.92%

B) P(2575 < X<3300) = 14.32%

C)  ( P X < 1225) = 4.55%

Step-by-step explanation:

Given data:

\mu = $2060

\sigma  = $495

a) P(x> 2575)

 1 - P(X<2575)

1 - P(\frac{x-\mu}{\sigma} < \frac{2575 - 2060}{495})

1 - P(z < \frac{515}{495})

1 - P( Z< 1.04)

1 - 0.8508

0.1492

14.92%

B) P(2575 < X<3300)

P(\frac{2575 - 2060}{495}

P (1.04 < z< 2.51)

P(z <2.51) - P(Z<1.04)

0.994 - 0.8508 = 0.1432 = 14.32%

C) ( P X < 1225)

P( \frac{x - \mu}{\sigma} < \frac{1225 - 2060}{495})

P( z < \frac{-835}{495}

P ( Z< -1.69)

= 0.0455 = 4.55%

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How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
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Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
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Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

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