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ycow [4]
3 years ago
13

James, a real estate appraiser is looking at homes in the 78681 zip code. He takes a random sample of 42 homes form the Austin M

LS (that's where Realtors go for information) and finds that 29 of them are listed as "eclectic" style. What is the 80% confidence interval for the proportion of homes that are "eclectic"?
Mathematics
1 answer:
vazorg [7]3 years ago
7 0

Answer:

80% confidence interval for the proportion of homes that are "eclectic" is [0.59 , 0.78].

Step-by-step explanation:

We are given that James takes a random sample of 42 homes form the Austin MLS (that's where Realtors go for information) and finds that 29 of them are listed as "eclectic" style.

Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;

                              P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of homes that are "eclectic" = \frac{29}{42} = 0.69

            n = sample of homes = 42

            \mu = population proportion of homes that are "eclectic"

<em>Here for constructing 80% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 80% confidence interval for the population proportion, p is ;</u>

P(-1.282 < N(0,1) < 1.282) = 0.80  {As the critical value of z at 10% level

                                                     of significance are -1.282 & 1.282}  

P(-1.282 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.282) = 0.80

P( -1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.80

P( \hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.80

<u>80% confidence interval for p</u> = [ \hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } },\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.69-1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } } , 0.69+1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } } ]

= [0.59 , 0.78]

Therefore, 80% confidence interval for the proportion of homes that are "eclectic" is [0.59 , 0.78].

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Step-by-step explanation:

For each frame, there are only two possible outcomes. Either there is a strike, or there is not. The probability of a strike happening in a frame is independent of other frames. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

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This means that p = 0.85

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This is P(X = 0) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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0.34% probability that she goes three consecutive frames without a​ strike.

​b) makes her first strike in the third ​frame?

No strike during the first two(with a 15% probability)

Strike during the third(85% probability). So

P = 0.15*0.15*0.85 = 0.0191

1.91% probability that she her first strike in the third ​frame

c) has at least one strike in the first three ​frames? ​

Either there are no strikes, or there is at least one strike. The sum of the probabilities of these events is 100%.

From a), 0.34% probability that she goes three consecutive frames without a​ strike.

100 - 0.34 = 99.66

99.66% probability that she has at least one strike in the first three ​frames.

d) bowls a perfect game​ (12 consecutive​ strikes)?

This is P(X = 12) when n = 12. So

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P(X = 12) = C_{12,12}.(0.85)^{12}.(0.15)^{0} = 0.1422

14.22% probability that she bowls a perfect game.

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