Question

In the western region of the state the times of all boys running in this event are normally distributed with standard deviation 12 seconds, but with mean 5 minutes 22 seconds. Find the proportion of boys from this region who qualify to run in this event in the state meet. (Hint: normalcdf)

Answer 1

Here is the full question.

The average finishing time among all high school boys in a particular track event in a certain state is 5 minutes 17 seconds. Times are normally distributed with standard deviation 12 seconds.

A.  The qualifying time in this event for participation in the state meet is to be set so that only the fastest 5% of all runners qualify. Find the qualifying time in seconds (round it to the closest second). (Hint: Convert minutes to seconds.)  

B. In the western region of the state the times of all boys running in this event are normally distributed with standard deviation 12 seconds, but with mean 5 minutes 22 seconds. Find the proportion of boys from this region who qualify to run in this event in the state meet. (Hint: normalcdf)

Answer:

a. x ≅ 337 seconds.

b. P(x > 337 )  = 0.1056

Step-by-step explanation:

A.

Given that ;

Mean $\mu =$ 5 minutes 17 seconds =( (60× 5)+17  ) seconds  = 317 seconds   ( since 60 seconds make 1 minute.

Standard deviation: $\sigma$ = 12 seconds.

Only the fastest 5% of all runners qualify

The objective is to determine the qualifying time in seconds

Let's look for the Z-score of 0.95;

The Z-score is 1.645 from the tables

$x= ( \sigma * z ) + \mu$

$x = ( 12 * 1.645 ) + 317 \\ \\x = 336.74$

x ≅ 337 seconds.

B. Given that the standard deviation = 12 seconds

Mean = 5 minutes 22 seconds = (5 × 60 + 22 )seconds = 322 seconds

he objective is  to find P(x > 337 )   i.e the proportion of boys from this region who qualify to run in this event in the state meet.

we are using command normalcdf   (SEE THE ATTACHED FILE BELOW FOR THE COMPUTATION)

we have  P(x > 337 )  = 0.1056