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masha68 [24]
3 years ago
14

Student A has a probability of 0.9 of solving a problem while student B has a probability of 0.6 of solving the problem. What is

the probability that both of them will solve the problem?
Mathematics
1 answer:
Vilka [71]3 years ago
6 0

answer

0.54

explanation

since student A and student B are independent of each other, we multiply their individual probabilities to get the probability of both solving the problem

A * B

= 0.9 * 0.6

= 0.54

there is a 0.54 probability that both will solve the problem

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Answer:

(a) Null Hypothesis, H_0 : p_1-p_2=0  or  p_1= p_2  

    Alternate Hypothesis, H_A : p_1-p_2\neq 0  or  p_1\neq p_2

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

<em />

<em>Let </em>p_1<em> = proportion of drivers that were in compliance initially</em>

p_2<em> = proportion of drivers that were in compliance one year later</em>

(a) <u>Null Hypothesis</u>, H_0 : p_1-p_2=0  or  p_1= p_2      {means that there is not any statistical difference in these two proportions measured initially and then one year later}

<u>Alternate Hypothesis</u>, H_A : p_1-p_2\neq 0  or  p_1\neq p_2     {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is <u>Two-sample z proportion statistics</u>;

                     T.S.  = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers in compliance initially = 98.9%

\hat p_2 = sample proportion of drivers in compliance one year later = 96.9%

n_1 = sample of drivers initially = 1,250

n_2 = sample of drivers one year later = 1,100

(b) So, <u><em>the test statistics</em></u>  =  \frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }  

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<em>Since our test statistics does not lies within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis.</em></u>

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

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