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Alex73 [517]
2 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]2 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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what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
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the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
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