Suppose you are interested in measuring the amount of time, on average, it takes

Question

Suppose you are interested in measuring the amount of time, on average, it takes you to make your communte to school. You've estimated that the average time is 38.15 minutes with a standard deviation of 7 minute. Over 12 random days, you measure your communte time. Assuming that your estimated parameters are correct, what is the probability that the average commute time over that 12 day period is between 35.76 and 37.57 minutes?

A. The sample mean will never fall in this range

B. 0.0913

C. 0.2686

D. 0.5055

E. 0.1006

Explanation / Answer

Solution:

Answer C:

conver x to z by

z=x-mean/sd/sqrt( n)

for x=35.76

mean=38.15

sd=7

n=12

z=35.76-38.15/7/sqrt(12)

=-1.182

For X=37.57

z=37.57-38.15/7/sqrt(12)

=-0;287

To find the probability of P (1.187<Z<0.287), we use the following formula:

P (1.187<Z<0.287 )=P ( Z<0.287 )P (Z<1.187 )

Step 3: P ( Z<0.287 ) can be found by using the following fomula.

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

After substituting a=0.287 we have:

P ( Z<0.287)=1P ( Z<0.287 )

P ( Z<0.287 ) can be found by using the following standard normal table.

From Standard Normal Table

We see that P ( Z<0.287 )=0.6141 so,

P ( Z<0.287)=1P ( Z<0.287 )=10.6141=0.3859

Step 4: P ( Z<1.187 ) can be found by using the following fomula.

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

After substituting a=1.187 we have:

P ( Z<1.187)=1P ( Z<1.187 )

P ( Z<1.187 ) can be found by using the following standard normal table.

From Standard Normal Table

We see that P ( Z<1.187 )=0.883 so,

P ( Z<1.187)=1P ( Z<1.187 )=10.883=0.117

At the end we have:

P (1.187<Z<0.287 )=0.2689

ANSWER:C. 0.2686

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