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Advocard [28]
3 years ago
9

A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken

if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within 0.01 with probability 0.95? (Round your answer up to the nearest whole number.)
Mathematics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

X=1 number of citrus trees that are infected with the citrus red mite.

n=5 random sample taken

\hat p=\frac{1}{5}=0.2 estimated proportion of citrus trees that are infected with the citrus red mite.

p true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56  

And rounded up we have that n=6147

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Multiply this distance by $0.89 per km.

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7nadin3 [17]
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83/10
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3 years ago
Five pounds is 2.268 times as heavy as one kilogram. How many times as heavy is two pounds than one kilogram?
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7 0
3 years ago
Suppose that the quantity supplied S and quantity demanded D of​ T-shirts at a concert are given by the following functions wher
harkovskaia [24]

Answer:

Step-by-step explanation:

The demand function is expressed as

D(p )=1150-50p

The supply function is expressed as S(p)=−200 + 40p

Where

p represents the price

A) The equilibrium price is the price at which the quantity supplied and the quantity demanded would be equal. Therefore

1150 - 50p = - 200 + 40p

40p + 50p = 1150 + 200

90p = 1350

p = 1350/90

p = 15

The equilibrium price is $15

b) For quantity demanded to be greater than quantity supplied, the price would be

1150 - 50p > - 200 + 40p

40p + 50p > 1150 + 200

90p > 1350

p > 1350/90

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The price would be greater than 15

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Base of rectangle = (6 - 2) / 2
= 2

The area of the first rectangle:

(4 - 2)f(4) = 2[4 + cos(4π)]


The area the second triangle:

(6 - 4)f(6) = 2[6 + cos(6π)]


Now just compute the two areas and combined them. That will give you the estimated under the curve.


To evaluate the midpoint of each rectangle, we take the midpoint of the base lengths of each rectangle. This midpoint is the x value. Then evaluate the function at that x value.


The midpoint of the first rectangle is x=3. Evaluate f(3).
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7 0
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