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GenaCL600 [577]
3 years ago
10

The incomes of all families in a particular suburb can be represented by a continuous random variable. It is known that the medi

an income for all families in this suburb is $60,000 and that 40% of all families in the suburb have incomes above $72,000.
a) For a randomly chosen family what is the probability that its income will be between $60000 and $72000.

b) If the distribution function for the income is known to be uniform what is the probability that a random chosen family has an income below $65000.
Business
1 answer:
liberstina [14]3 years ago
7 0

Answer:

a) 0.10 or 10%

b) 0.5417 or 54.17%

Explanation:

a) The median income of $60,000 is at the 50th percentile of the distribution. If 40% if incomes are above $72,000, then an income of $72,000 is at the 60th percentile of the distribution. Therefore, the probability that a family's income will be between $60,000 and $72,000 is:

P( \$60,000 \leq X \leq \$72,000)=0.6-0.5\\P( \$60,000 \leq X \leq \$72,000)=0.1 = 10\%

b) If the distribution is known to be uniform, the probability that a random chosen family has an income below $65,000 is:

P( X \leq \$65,000)=0.5+\frac{65,000-60,000}{72,000-60,000}*0.1\\ P( X \leq \$65,000)=0.5417 =54.17\%

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Explanation:

First we construct the equation system:

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Now we clear one and replace:

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And we can solve for type B:

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And now we can solve for quantity of A as well:

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<u>Finally we can check the answer if it is correct:</u>

50 x 5.9 + 94 X 4.75 =

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