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STALIN [3.7K]
3 years ago
5

A particular moth can infest avocado trees and potentially damage the avocado fruit. A farmer plans to investigate the number of

damaged avocado fruit on moth-infested avocado trees of a certain variety. A recent article states that the population distribution of the number of damaged avocado fruit on moth-infested avocado trees of this variety is symmetric with mean of 6.4 and standard deviation of 1.9.
Required:

Compare the shapes of the sampling distributions of the sample mean number of damaged avocado fruit for random samples of 6 moth-infested trees and for random samples of 90 moth-infested trees from the population. Explain your answer.
Mathematics
1 answer:
rewona [7]3 years ago
6 0

Answer:

By the Central Limit Theorem, both distributions are normal, with mean of 6.4. For samples of size 6, the standard deviation of the the sampling distributions of the sample mean number of damaged avocado fruit is of 0.78, while for samples of 90, it is of 0.2.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

A recent article states that the population distribution of the number of damaged avocado fruit on moth-infested avocado trees of this variety is symmetric with mean of 6.4 and standard deviation of 1.9.

So normally distributed, with \mu = 6.4, \sigma = 1.9

Samples of 6:

This means that n = 6, s = \frac{1.9}{\sqrt{6}} = 0.78

The mean is the same.

Samples of 90:

This means that n = 90, s = \frac{1.9}{\sqrt{90}} = 0.2

The mean is the same.

By the Central Limit Theorem, both distributions are normal, with mean of 6.4. For samples of size 6, the standard deviation of the the sampling distributions of the sample mean number of damaged avocado fruit is of 0.78, while for samples of 90, it is of 0.2.

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A line passes through the points ( – 4, – 2) and ( - 1, - 2). Determine the slope of the line.
Veseljchak [2.6K]

Answer: -4/3

Step-by-step explanation:

Formula to find a slope of two given points is

y(sub2) - y(sub1) / x(sub2) - x(sub1)

Plug the values in to get the answer.

-2 - 2 / -1 - (-4)

-4/3

4 0
4 years ago
Find the expansions of sin⁡5θ and cos⁡5θ in terms of sin⁡θ and cos⁡θ.
qwelly [4]
\bf \textit{quad-angle identities}\\\\\\

cos(4\theta)=8cos^4(\theta)-8cos^2(\theta)+1
\\\\\\
sin(4\theta)=
\begin{cases}
8sin(\theta)cos^3(\theta)-4sin(\theta)cos(\theta)\\\\
4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)
\end{cases}\\\\
-----------------------------\\\\


\bf sin(5\theta)\iff sin(4\theta + \theta)
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sin(4\theta)cos(\theta)+cos(4\theta)sin(\theta)
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[4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)]cos(\theta)\\\\ + [8cos^4(\theta)-8cos^2(\theta)+1]sin(\theta)

\bf -----------------------------\\\\
cos(5\theta)\iff cos(4\theta + \theta)
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cos(4\theta)cos(\theta)-sin(4\theta)sin(\theta)\\\\\\\
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4 0
3 years ago
Help please anything
Nat2105 [25]

Answer:

36 ft

Step-by-step explanation:

The ratio of the lengths of the sides is the same as the ratio of the perimeters.

Just like one side (15 ft) of triangle ABC is 3 times as long as the corresponding side (5 ft) of triangle DEF, the perimeter of triangle ABC is also 3 times the perimeter of triangle DEF.

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7 0
3 years ago
A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta
Akimi4 [234]

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true 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 women who find success with this new treatment = \frac{7}{14} = 0.50

          n = sample of women = 14

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the true proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                            level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u />

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } , 0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } ]

 = [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

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Answer:

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