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Fittoniya [83]
3 years ago
9

Suppose the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

ounces. suppose carl bags his potatoes in randomly selected groups of 6. what percentage of these bags should have a mean potato weight between 7.5 and 8.5 ounces?
Mathematics
1 answer:
denis23 [38]3 years ago
6 0
Given that <span>the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.

The probability of a normally distributed data between two values (a, b) is given by:

P(a\ \textless \  X\ \textless \  b)=P\left(z\ \textless \  \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left(z\ \textless \  \frac{a-\mu}{\sigma/\sqrt{n}} \right) \\  \\ =P(7.5\ \textless \ X\ \textless \ 8.5)=P\left(z\ \textless \  \frac{8.5-8.0}{1.1/\sqrt{6}} \right)-P\left(z\ \textless \  \frac{7.5-8.0}{1.1/\sqrt{6}} \right) \\  \\ =P(z\ \textless \ 1.113)-P(-1.113)=0.8672-0.1328=0.7344</span>
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three friends are in the same algebra class. Their scores on a recent test are three consecutive odd intergers whose sum is 279.
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X= 1st integer
x+2= 2nd integer
x+4= 3rd integer

Add the integers together

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combine like terms
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Substitute x=91 to find 2nd & 3rd integers

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ANSWER: The three test scores are 91, 93 and 95.

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3 years ago
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When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the appr
marin [14]

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96.42% probability that the number of tails is between 40 and 60.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

100 tosses, so n = 100

Two outcomes, both equally as likely. So p = \frac{1}{2} = 0.5

So

E(X) = np = 100*0.5 = 50

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5

Estimate the approximate probability that the number of tails is between 40 and 60.

Using continuity correction.

P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)

This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So

X = 60.5

Z = \frac{X - \mu}{\sigma}

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X = 39.5

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Z = \frac{39.5 - 50}{5}

Z = -2.1

Z = -2.1 has a pvalue of 0.0179

0.9821 - 0.0179 = 0.9642

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