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Westkost [7]
3 years ago
9

Use the sample data and confidence level to construct the confidence interval estimate of the population proportion p. n equals

550 comma x equals 440 comma 95 % confidence nothingless thanpless than nothing ​(Round to three decimal places as​needed.)
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

(0.767,0.833)

Step-by-step explanation:

The 95% confidence interval for population proportion p can be computed as

p-z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} }

The z-value associated with 95% confidence level is 1.96.

whereas p=x/n

We are given that x=440 and n=550.

p=440/550=0.8

0.8-1.96\sqrt{\frac{0.8(0.2)}{550} }

0.8-1.96\sqrt{\frac{0.16}{550} }

0.8-1.96\sqrt{0.00029 }

0.8-1.96(0.01706)

0.8-0.03343

0.76657

Thus, the required confidence interval is

0.767<P<0.833  (rounded to 3 decimal places)

Hence, we are 95% confident that our true population proportion will lie in the interval (0.767,0.833)

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One serving of muffins takes 2/3 cup of muffin mix. how many cups of muffin mix would it take to make 12 servings of muffins
Leno4ka [110]

1 serving = 2/3 cup

12 serving = 2/3 x 12 = 8 cups

Answer: 12 servings of muffin would need <u>8 cups</u> of the muffin mix.

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Martin has a business washing cars. Last year he washed 20 cars a week. This year, he wants to increase his business to 1,200 ca
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3 years ago
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
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timama [110]

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Step-by-step explanation:

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Where

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σ = standard deviation

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z = (40000 - 47500)/5000 = - 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.067

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