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22:06 10月20日周日 16%16 \% HW4 HW4 开始 播入 页面 审阅 衧图 效室 R. WPS AI Times 小四 A∗A^{*} $A^...
Oct 20, 2024
Solution by Steps
step 1
To develop the empirical discrete probability distribution for x x , we first need to calculate the total number of graduates surveyed, which is N=506+390+310+218+576=2000 N = 506 + 390 + 310 + 218 + 576 = 2000
step 2
Next, we calculate the probability for each value of x x (the number of years with the current employer) using the formula: P(x)=Number of graduates with x yearsN P(x) = \frac{\text{Number of graduates with } x \text{ years}}{N} . The probabilities are: - For x=1 x = 1 : P(1)=5062000=0.253 P(1) = \frac{506}{2000} = 0.253 - For x=2 x = 2 : P(2)=3902000=0.195 P(2) = \frac{390}{2000} = 0.195 - For x=3 x = 3 : P(3)=3102000=0.155 P(3) = \frac{310}{2000} = 0.155 - For x=4 x = 4 : P(4)=2182000=0.109 P(4) = \frac{218}{2000} = 0.109 - For x=5 x = 5 : P(5)=5762000=0.288 P(5) = \frac{576}{2000} = 0.288
step 3
To verify that this is a valid discrete probability distribution, we check that all probabilities are between 0 and 1 and that they sum to 1: P(1)+P(2)+P(3)+P(4)+P(5)=0.253+0.195+0.155+0.109+0.288=1.000 P(1) + P(2) + P(3) + P(4) + P(5) = 0.253 + 0.195 + 0.155 + 0.109 + 0.288 = 1.000
step 4
To find the probability that a respondent has been at their current place of employment for more than 3 years, we calculate: P(x > 3) = P(4) + P(5) = 0.109 + 0.288 = 0.397
Answer
The empirical discrete probability distribution is valid, and the probability that a respondent has been employed for more than 3 years is 0.397.
Key Concept
Empirical discrete probability distribution and its validation.
Explanation
The solution involves calculating probabilities based on observed frequencies and ensuring they meet the criteria for a valid distribution. The final probability indicates the likelihood of longer employment durations.
Solution by Steps
step 1
To compute the hypergeometric probabilities, we use the hypergeometric probability formula: P(X=x)=(rx)(N−rn−x)(Nn) P(X = x) = \frac{{\binom{r}{x} \binom{N-r}{n-x}}}{{\binom{N}{n}}} where N N is the population size, r r is the number of successes in the population, n n is the number of draws, and x x is the number of observed successes
step 2
For n=4,x=1 n = 4, x = 1 : We have N=10 N = 10 and r=3 r = 3 . Thus, P(X=1)=(31)(73)(104)=3⋅35210=105210=0.5 P(X = 1) = \frac{{\binom{3}{1} \binom{7}{3}}}{{\binom{10}{4}}} = \frac{3 \cdot 35}{210} = \frac{105}{210} = 0.5
step 3
For n=2,x=2 n = 2, x = 2 : P(X=2)=(32)(70)(102)=3⋅145=345=0.0667 P(X = 2) = \frac{{\binom{3}{2} \binom{7}{0}}}{{\binom{10}{2}}} = \frac{3 \cdot 1}{45} = \frac{3}{45} = 0.0667
step 4
For n=2,x=0 n = 2, x = 0 : P(X=0)=(30)(72)(102)=1⋅2145=2145=0.4667 P(X = 0) = \frac{{\binom{3}{0} \binom{7}{2}}}{{\binom{10}{2}}} = \frac{1 \cdot 21}{45} = \frac{21}{45} = 0.4667
step 5
For n=4,x=2 n = 4, x = 2 : P(X=2)=(32)(72)(104)=3⋅21210=63210=0.3 P(X = 2) = \frac{{\binom{3}{2} \binom{7}{2}}}{{\binom{10}{4}}} = \frac{3 \cdot 21}{210} = \frac{63}{210} = 0.3
step 6
For n=4,x=4 n = 4, x = 4 : P(X=4)=(34)(70)(104)=0 P(X = 4) = \frac{{\binom{3}{4} \binom{7}{0}}}{{\binom{10}{4}}} = 0 (since (34)=0 \binom{3}{4} = 0 )
Answer
A
(for n=4,x=1 n = 4, x = 1 )
B
(for n=2,x=2 n = 2, x = 2 )
C
(for n=2,x=0 n = 2, x = 0 )
D
(for n=4,x=2 n = 4, x = 2 )
E
(for n=4,x=4 n = 4, x = 4 )
Key Concept
Hypergeometric Distribution
Explanation
The hypergeometric distribution models the probability of x x successes in n n draws without replacement from a finite population.
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