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-BARSIC- [3]
3 years ago
5

The probability that will rain tomorrow 0.85. what is the probability that will not rain tomorrow?

Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer:

probability that will not rain tomorrow

P(nR) = 1-0.85

= 0.15

is the answer

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Answer: 20%

Step-by-step explanation:

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100  x

100*5/1 = 20%

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Cheryl has a cake recipe which calls for 3/4 a cup of flour. The only measuring cup she has is 1/8 a cup. How many cups should s
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In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. This is the 14
lbvjy [14]

Answer:

We are confident at 99% that the difference between the two proportions is between 0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420

Step-by-step explanation:

Part a

Data given and notation  

X_{D}=3266 represent the number people registered as Democrats

X_{R}=2137 represent the number of people registered as Republicans

n=7525 sampleselcted

\hat p_{D}=\frac{3266}{7525}=0.434 represent the proportion of people registered as Democrats

\hat p_{R}=\frac{2137}{7525}=0.284 represent the proportion of people registered as Republicans

The standard error is given by this formula:

SE=\sqrt{\frac{\hat p_D (1-\hat p_D)}{n_{D}}+\frac{\hat p_R (1-\hat p_R)}{n_{R}}}

And the standard error estimated given by the problem is 0.008

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion of Democrats that approve of the way the California Legislature is handling its job  

\hat p_A =\frac{1894}{3266}=0.580 represent the estimated proportion of Democrats that approve of the way the California Legislature is handling its job  

n_A=3266 is the sample size for Democrats

p_B represent the real population proportion of Republicans that approve of the way the California Legislature is handling its job  

\hat p_B =\frac{385}{2137}=0.180 represent the estimated proportion of Republicans that approve of the way the California Legislature is handling its job

n_B=2137 is the sample for Republicans

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

(0.580-0.180) - 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.380  

(0.580-0.180) + 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.420  

And the 99% confidence interval would be given (0.380;0.420).  

We are confident at 99% that the difference between the two proportions is between 0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420

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No link or bot answer the question
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