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Floretta's points per basketball game are normally distributed with a standard deviation of 4 points. If Floretta scores 10
points, and the Z-score of this value is - 4, then what is her mean points in a game? Do not include the units in your answer.
For example, if you found that the mean is 33 points, you would enter 33.

Sagot :

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.

jacob193

Answer:

[tex]26[/tex] points.

Step-by-step explanation:

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

[tex]\begin{aligned} z&= \frac{x - \mu}{\sigma}\end{aligned}[/tex].

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

[tex]\begin{aligned} x - \mu = \sigma\, z\end{aligned}[/tex].

[tex]-\mu = (-x) + \sigma\, z[/tex].

[tex]\begin{aligned}\mu = x - \sigma\, z\end{aligned}[/tex].

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

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

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