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Vinvika [58]
3 years ago
12

A pet store contains 35 light green parakeets (14 females and 21 males) and 44 sky blue parakeets (28 females and 16 males). You

randomly choose one of the parakeets. What is the probability that it is a female or a sky blue parakeet?
Mathematics
1 answer:
snow_tiger [21]3 years ago
7 0
You can use this formula <span>P(AorB) = P(A) + P(B) - P(AandB) 

Given:
35 LG (14 F & 21 M)
44 SB (28 F & 16 M)

Req:
- the probability that it is a female (F) or a sky blue (SB)

Sol:
</span>P(F or SB) = P(F) + P(SB) - P(F and SB) 
                 = [(14 F + 28 F)/(35 + 44)] + [(44 SB)/(35 + 44)] - [(28 F)/(35 + 44)]
                 = 53.16 + 55.70 - 35.44
                 = 73.42%

You have to deduct 28 female parakeets from 44 sky blue parakeets because the 28 parakeets are already accounted for in the female parakeets. You can also think of how many ways you can choose a female parakeet and a sky blue parakeet. Add all female parakeets (14 + 28) = 42. Sky blue parakeet equaled to 44. Minus the 28 female parakeets included in the sky blue parakeet to avoid double counting. 42 + 44 - 28 = 58 divided by 79 (35 + 44) total parakeets = 73.42%


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The U.S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is regi
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We conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

Step-by-step explanation:

We are given that 513 employed persons and 604 unemployed persons are independently and randomly selected, and that 287 of the employed persons and 280 of the unemployed persons have registered to vote.

Let p_1 = <u><em>percentage of employed workers who have registered to vote.</em></u>

p_2 = <u><em>percentage of unemployed workers who have registered to vote.</em></u>

So, Null Hypothesis, H_0 : p_1\leq p_2      {means that the percentage of employed workers who have registered to vote does not exceeds the percentage of unemployed workers who have registered to vote}

Alternate Hypothesis, H_A : p_1>p_2     {means that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote}

The test statistics that would be used here <u>Two-sample z test for proportions;</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 employed workers who have registered to vote = \frac{287}{513} = 0.56

\hat p_2 = sample proportion of unemployed workers who have registered to vote = \frac{280}{604} = 0.46

n_1 = sample of employed persons = 513

n_2 = sample of unemployed persons = 604

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                                       =  3.349

The value of z test statistics is 3.349.

<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>

Since our test statistic is more than the critical value of z as 3.349 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

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