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Ahat [919]
3 years ago
9

How do I make this bell curve

Mathematics
1 answer:
Zanzabum3 years ago
8 0

The mean you found is correct, but the standard deviation is not. Recall that the standard deviation \sigma_s (s for sample) of n points is given by

\sigma_s=\sqrt{\dfrac1{n-1}\displaystyle\sum_{1\le i\le n}(x_i-\bar x)^2}

where n=10 is the sample size, \bar x=3740 is the sample mean, and x_i are the prices listed in the circled column. So

\sigma_s=\sqrt{\dfrac{(3640-3740)^2+(7595-3740)^2+\cdots+(3390-3740)^2}{10-1}}

\implies\sigma_s\approx1443.98\approx1444

I can't tell if you need to provide any more info beyond this, but given there's a plot of a generalized bell curve, I think you're also supposed to label the plot.

At the center of the bell-shaped/normal distribution is the mean. Notice there are three tick marks to either side of the mean - these are probably supposed to represent prices that fall exactly 1, 2, and 3 standard deviations from the mean. These are, from left to right,

\bar x-3\sigma_s\approx3740-3(1444)=-592

\bar x-2\sigma_s\approx3740-2(1444)=852

\bar x-\sigma_s\approx2296

\bar x+\sigma_s\approx5184

\bar x+2\sigma_s\approx6628

\bar x+3\sigma_s\approx8072

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3 years ago
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A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a
Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y  grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\

\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)  [as x+y=14]

\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\

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and y = 14-x= 14-10.9375=3.0625 grams.

Hence, 10.9375 grams of 82% gold and 3.0625  grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y  grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x)  \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\

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and y = 14-x= 14-12.28=1.72 grams.

Hence, 12.28 grams of 82% gold and 1.72  grams of 50% gold added to make 14 grams of 75% gold.

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Answer:

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(2) 3(6 +7n) = 87 . . . . . . add 5

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