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kodGreya [7K]
2 years ago
15

Floretta's points per basketball game are normally distributed with a standard deviation of 4 points. If Floretta scores 10

Mathematics
2 answers:
Alja [10]2 years ago
7 0

Answer:

26 points.

Step-by-step explanation:

Let X denote a normal random variable with mean \mu and standard deviation \sigma. (That is: X \sim {\rm N}(\mu,\, \sigma).) By definition, the z-score of an observation with value X = x would be:

\begin{aligned} z&= \frac{x - \mu}{\sigma}\end{aligned}.

In this question, the value of \sigma is given. Also given are the value of the observation x and the corresponding z-score, z\!. Rearrange the \! z-score definition z = (x - \mu) / \sigma to find an expression for \mu:

\begin{aligned} x - \mu = \sigma\, z\end{aligned}.

-\mu = (-x) + \sigma\, z.

\begin{aligned}\mu = x - \sigma\, z\end{aligned}.

Substitute in the value of x, \sigma, and z to find the value of \mu, the mean of this normal random variable:

\begin{aligned}\mu &= x - \sigma\, z \\ &= 10 - (-16) \\ &= 26\end{aligned}.

irga5000 [103]2 years ago
3 0

Answer:

26

Step-by-step explanation:

We can work backwards using the z-score formula to find the mean. The problem gives us the values for z, x and σ. So, let's substitute these numbers back into the formula:

z−4−16−2626=x−μσ=10−μ4=10−μ=−μ=μ

We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=10 is four standard deviations to the left of the mean, and each standard deviation is 4. So four standard deviations is (−4)(4)=−16 points. So, now we know that 10 is 16 units to the left of the mean. (In other words, the mean is 16 units to the right of x=10.) So the mean is 10+16=26.

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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

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x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

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\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

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When Joe arrives:

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\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

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\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

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L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

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