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Phoenix [80]
3 years ago
11

In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre

, with a standard deviation of 23.5 bushels of corn. If a study included 1100 acres, about how many would be expected to yield more than 190 bushels of corn per acre?
A. 639 acresB. 461 acresC. 419 acresD. 503 acres
Mathematics
1 answer:
RideAnS [48]3 years ago
8 0

Answer: B. 461 acres

Step-by-step explanation:

Given : In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn.

i.e. \mu=185.2\ \ , \ \sigma=23.5

Let x denotes the amount of corn yield.

Now, the probability that the amount of corn yield is more than 190 bushels of corn per acre.

P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})

[Formula : z=\dfrac{x-\mu}{\sigma}]

=P(z>0.2043)=1-P(z  [∵ P(Z>z)=1-P(Z<z)]

1-0.5809405   [using z-value calculator or table]

=0.4190595

Now, If a study included 1100 acres then the expected number to yield more than 190 bushels of corn per acre :-

0.4190595\times1100=460.96545\approx461\text{ acres}

hence, the correct answer is B. 461 acres .

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Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

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\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})

Compute the sample sizes as follows:

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

       n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}

          =\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

\hat p_{1}=0.45\\\hat p_{2}=0.58

Compute the sample size as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}

       n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}

          =\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666

Thus, the sample sizes are 6666.

7 0
3 years ago
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Step-by-step explanation:

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Step-by-step explanation:

x^2+18=-9x

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Simplify:

x^2+9x+18=0

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x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

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