Question

Kathy and her brother Clay recently ran in a local marathon. The distribution of finishing time for women was approximately normal with mean 259 minutes and standard deviation 32 minutes. The distribution of finishing time for men was approximately normal with mean 242 minutes and standard deviation 29 minutes. What is the probability of the menand women running a marathon time of less than 2 hours?

Answer 1

Answer:

P (X<120)=0

Step-by-step explanation:

We can compute the probability for women:

μ=259  and  σ=32

P ( X<120 )=P ( X−μ<120-259 )=P ((X−μ)/σ<(120−259)/32)

Since (x−μ)/σ =Z

and  

$Z=(120-259)/32=-4.34$

we have:

$P(X<120)=P (Z<-4.34)$

Use the standard normal table to find this area.

$P(Z<-a)=1-P(Z<a)$

$P(Z<-4.34)=1-P(Z<4.34)$

Now we can find  P ( Z<4.34 )  by using the standard normal table which is 1:

$P(Z<-4.34)=1-P(Z<4.34 )=1-1=0$

Therefore: P (X<120)=0

For men

μ=242  and  σ=29

P ( X<120 )=P ( X−μ<120-242 )=P ((X-μ)/σ <(120-242)/29  )

Since (x-μ)/σ=Z

and  

$Z=(120-242)/29=-4.21$

we have:

$P(X<120)=P (Z<-4.21)$

Use the standard normal table to find this area.

$P(Z<-a)=1-P(Z<a)$

$P(Z<-4.21)=1-P(Z<4.21)$

Now we can find  P ( Z<4.21 )  by using the standard normal table which is 1:

$P(Z<-4.21)=1-P(Z<4.21)=1-1=0$

Therefore: P (X<120)=0

Answer 2

Using the Zscore principle, the probability of men and women finishing in less than two hours is 0

Recall :

• Zscore = (x - mean) ÷ standard deviation
• x = score

For men :

• x = 2 hours = 120 minutes

Zscore = (120 - 242) / 29 = - 4.206

P(Z < -4.206) = 0.00001

For women :

• x = 2 hours = 120 minutes

Zscore = (120 - 259) / 32 = -4.34

P(Z < -4.34) = 0.00001

The probability of Men and Women time being below 2 hours :

• P(Z < -4.34) - P(Z < -4.206)

P(Z < -4.34) - P(Z < -4.206) = 0.00001 - 0.00001 = 0

Hence, the probability of men and women finishing in less than two hours is 0

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