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Fynjy0 [20]
3 years ago
6

1 point) As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distri

buted with a mean of 61 minutes and a standard deviation of 9 minutes. Let X denote finishing time for the finishers. a) The distribution of the variable X has mean 61 and standard deviation 9 . b) The distribution of the standardized variable Z has mean and standard deviation
Mathematics
1 answer:
BigorU [14]3 years ago
8 0

Answer:

(a) E (X) = 61 and SD (X) = 9

(b) E (Z) = 0 and SD (Z) = 1

Step-by-step explanation:

The time of the finishers in the New York City 10 km run are normally distributed with a mean,<em>μ</em> = 61 minutes and a standard deviation, <em>σ</em> = 9 minutes.

(a)

The random variable <em>X</em> is defined as the finishing time for the finishers.

Then the expected value of <em>X</em> is:

<em>E </em>(<em>X</em>) = 61 minutes

The variance of the random variable <em>X</em> is:

<em>V</em> (<em>X</em>) = (9 minutes)²

Then the standard deviation of the random variable <em>X</em> is:

<em>SD</em> (<em>X</em>) = 9 minutes

(b)

The random variable <em>Z</em> is the standardized form of the random variable <em>X</em>.

It is defined as:Z=\frac{X-\mu}{\sigma}

Compute the expected value of <em>Z</em> as follows:

E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

The mean of <em>Z</em> is 0.

Compute the variance of <em>Z</em> as follows:

V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1

The variance of <em>Z</em> is 1.

So the standard deviation is 1.

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