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Bas_tet [7]
3 years ago
11

In a chemical process, the amount of a certain type of impurity in the output is difficult to control and is thus a random varia

ble. It is known that the standard deviation of the impurity is 0.0015 following an approximately normal distribution. If a random sample of 75 outputs has an average impurity of 0.310, find a 95% confidence interval for the mean amount of the impurity.
Chemistry
2 answers:
Julli [10]3 years ago
8 0

Answer:

+/- 0.00033

Explanation:

For a 95 % confidence interval the range is given by

+/- Z * s/sqrt(n)

where Z value is 1.916 for a 95% confidence interval

Therefore the interval is

s is the standard deviation in this case 0.0015

n is 75 the number of outposts

calculating,

+/- 1.916 * 0.0015/sqrt(75) = +/- 0.00033

strojnjashka [21]3 years ago
5 0

Answer:

\\ (Lower\;value =0.3097 ; Upper\;value = 0.3103).

We can also represent it as \\ P(0.3097 \leq \mu \leq 0.3103) = 0.95.

Explanation:

This is a case of finding <em>two values</em> that define an <em>interval</em> in which is included the <em>population mean for the impurity</em> in the chemical process with a probability of 95%.

We have full information to this respect to solve the question:

  1. The <em>population standard deviation</em>, which is \\ \sigma = 0.0015.
  2. The <em>sample mean</em>, which is \\ \bar{x} = 0.310.
  3. The <em>sample size</em> \\ n = 75.

That is, we can find the 95% confidence interval for a <em>given population standard deviation</em>.

The formula for finding these <em>two values, </em>in these conditions,<em> </em>is as follows:

\\ Lower\;value = \bar{x} - 1.96\frac{\sigma}{\sqrt{n}}.

\\ Upper\;value = \bar{x} + 1.96\frac{\sigma}{\sqrt{n}}.

We have already confirmed that we have all these values: the population standard deviation, the sample mean and the sample size.

The number 1.96 corresponds to a <em>z-score</em> that is<em> 1.96</em> times the standard deviation from the mean, and represents the <em>confidence coefficient</em> and depends on the <em>confidence level</em>, which is in this case of 5% (or 0.05). A confidence level of 5% determines this 95 % confidence interval.

Notice that we are going to use a mean obtained from a sample of 75 elements \\ \bar{x} = 0.310, which represents a <em>sample mean</em>.

As a result, in order to find the two values that define the <em>95% confidence interval, </em>we can proceed as follows:

Lower value of the 95% confidence interval

\\ Lower\;value = \bar{x} - 1.96\frac{\sigma}{\sqrt{n}}.

\\ Lower\;value = 0.310 - 1.96\frac{0.0015}{\sqrt{75}}.

\\ Lower\;value =0.3097.

Upper value of the 95% confidence interval

\\ Upper\;value = \bar{x} + 1.96\frac{\sigma}{\sqrt{n}}.

\\ Upper\;value = 0.310 + 1.96\frac{0.0015}{\sqrt{75}}.

\\ Upper\;value = 0.3103.

Thus, the 95% confidence interval, or the interval for which there is a probability of 95% to find the population mean for this certain type of impurity:

\\ Lower\;value =0.3097.

\\ Upper\;value = 0.3103.

That is:

\\ P(0.3097 \leq \mu \leq 0.3103) = 0.95.

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102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

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From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

<u>Method 1: using the </u><u>mass of glucose</u>

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= 200 ÷180

= \frac{10}{9} mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

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Mass of ethanol

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= 102 g (3 s.f.)

<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>

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= \frac{977}{440} mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= \frac{977}{440} mol

Mass of ethanol formed

= \frac{977}{440} \times[2(12)+6+16]

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

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