Question

As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distributed 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 ______ and standard deviation _______ .b) The distribution of the standardized variable z has mean ________ and standard deviation ________.c) The percentage of finishers with times between 50 and 70 minutes is equal to the area under the standard normal curve between _____ and _____.d) The percentage of finishers with times exceeding 81 minutes is equal to the area under the standard normal curve that lies to the ______ of _____ .

Answer 1

For a) 61 minutes and 9 minutes

b) 0 and 1

c) -1.222 and 1

d) Right and 2.222

Explanation:

Let's start defining the random variable.

X : ''Times of the finishers in the New York City 10 km run ''

This random variable is normally distributed ⇒

X ~ N (μ,σ)

Where μ is the mean and σ is the standard deviation.

The mean and the standard deviation are the two parameters that define the normal distribution.

In the exercise, the mean is 61 minutes and the standard deviation is 9 minutes.

For a) The distribution of the variable x has mean 61 minutes and standard deviation 9 minutes.

If we want to calculate probabilities for the random variable X, we need to work with the standardized normal random variable.

This variable ''Z'' is normally distributed with a mean of 0 and a standard deviation of 1 ⇒

Z ~ N (0,1)

For b) The distribution of the standardized variable z has mean 0 and standard deviation 1.

To calculate the probability
`P(50<X<70)` we can standardize the variable to obtain the variable Z by subtracting the mean and dividing by the standard deviation :

P (50<X<70) = P [(50-μ)/σ < (X-μ)/σ < (70-μ)/σ]

Where (X-μ)/σ = Z We know that the mean is 61 minutes and the standard deviation is 9 minutes ⇒

`P(50<X<70)=P((50-61)/(9)<Z<(70-61)/(9))`

`P(50<X<70)=P(-1.222<Z<1)`

Where
`P(-1.222<Z<1)` is equal to the area under the standard normal curve between - 1.222 and 1

For c) The percentage of finishers with times between 50 and 70 minutes is equal to the area under the standard normal curve between -1.222 and 1

Finally, If we want the percentage of finishers with times exceeding 81 minutes we want the following probability :

`P(X>81)`

Standardizing this :

`P(X>81)=P((X-61)/(9)>(81-61)/(9))=P(Z>2.222)`

Where
`P(Z>2.222)` is equal to the area under the standard normal curve that lies to the right of 2.222

For d) The percentage of finishers with times exceeding 81 minutes is equal to the area under the standard normal curve that lies to the right of 2.222

Answer 2

As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distributed 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 ____0____ and standard deviation ____1____.c) The percentage of finishers with times between 50 and 70 minutes is equal to the area under the standard normal curve between __-1.22___ and __1___.d) The percentage of finishers with times exceeding 81 minutes is equal to the area under the standard normal curve that lies to the __right of 2.22____ .

Explanation:

a)- Mean = 61 and Standard Deviation = 9

b)- Mean = 0 and Standard Deviation = 1

c)- Using formula:
`z=(x-\mu)/(\sigma)`

`(50-61)/(9)\ \text{and} \ (70-61)/(9)`

⇒ z = -1.22 and z = 1

(d)
`z=(81-61)/(9)=2.22`

Thus, Area under Standard curve lies on the right of 2.22