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Over [174]
1 year ago
8

10.5% of the population is 65 or older. Find the probability that the following number of persons selected at random from 25 peo

ple are 65 or older.The probability that at most 2 are 65 or older is(Round to three decimal places as needed)
Mathematics
1 answer:
julia-pushkina [17]1 year ago
8 0

Given:

The sample size n=25

Probability of population is 65 or older is 10.5%.

This date follows the binomial distribution,

\begin{gathered} n=25,p=0.105 \\ X\rightarrow B(n=25,p=0.105) \end{gathered}

To find the probability that at most 2 are 65 or older,

\begin{gathered} P(X=x)=^nC_x(p)^x(1-p)^{n-x} \\ P(0\leq X\leq2)=P(X=0)+P(X=1)+P(X=2) \\ =^{25}C_0(0.105)^0(1-0.105)^{25-0}+^{25}C_1(0.105)^1(1-0.105)^{25-1}+^{25}C_2(0.105)^2(1-0.105)^{25-2} \\ =0.0625+0.1832+0.2579 \\ =0.5036 \\ \approx0.504 \end{gathered}

Answer: Probability is 0.504 .

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P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

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c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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answer
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Read 2 more answers
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