Circle Z is shown. Line segments Z T, Z U, and Z V are radii of the circle. Line segment U X goes through point X and is a diame
2 answers:
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Answer:
(a) Probability mass function
P(X=0) = 0.0602
P(X=1) = 0.0908
P(X=2) = 0.1704
P(X=3) = 0.2055
P(X=4) = 0.1285
P(X=5) = 0.1550
P(X=6) = 0.1427
P(X=7) = 0.0390
P(X=8) = 0.0147
NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.
(b) Cumulative distribution function of X
F(X=0) = 0.0602
F(X=1) = 0.1510
F(X=2) = 0.3214
F(X=3) = 0.5269
F(X=4) = 0.6554
F(X=5) = 0.8104
F(X=6) = 0.9531
F(X=7) = 0.9921
F(X=8) = 1.0068
Step-by-step explanation:
Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).
<u>For X=0</u>
This happens when every couple and the singles are on time (ot).
<u>For X=1</u>
This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:
<u>For X=2</u>
This happens when
1) Only one of the three couples is late, and the others cooples and singles are on time.
2) When both singles are late , and the couples are on time.
<u>For X=3</u>
This happens when
1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.
<u>For X=4</u>
This happens when
1) Only two couples are late. There are 3 combinations of these.
2) Only one couple and both singles are late. Only one combination of these situation.
<u>For X=5</u>
This happens when
1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.
<u>For X=6</u>
This happens when
1) Only the three couples are late (1 combination)
2) Only two couples (3 combinations) and one single (2 combinations) are late
<u>For X=7</u>
This happens when
1) Only one of the singles is on time (2 combinations)
<u>For X=8</u>
This happens when everybody is late
Answer:
There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, <em>assuming a significance level 0.1</em>
Step-by-step explanation:
Let p1 be the proportion of West University students who involved in a car accident within the past year
Let p2 be the proportion of East University students who involved in a car accident within the past year
Then
p1=p2
p1≠p2
The formula for the test statistic is given as:
z= where
- p1 is the <em>sample</em> proportion of West University students who involved in a car accident within the past year (0.15)
- p2 is the <em>sample</em> proportion of East University students who involved in a car accident within the past year (0.12)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the students from West University (100)
- n2 is the sample size ofthe students from East University (100)
Then we have z= ≈ 0.6208
Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.
<em>Assuming significance level 0.1</em>, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance