g 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

Complete 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)

a

    The qualifying time is  

b

  The proportion of boy that qualify id 0.0197 of all the boys from west region that participated in the event

Step-by-step explanation:

Considering  a

   The mean is  \

     The standard deviation is

     The percentage of success is  5% =0.05

Now since the times is normally distributed the critical value of success obtained from the standardized normal distribution table is

So the qualifying time is mathematically represented as

substituting values

Considering  b

   The means is  

Now from a the qualifying time is  

Now the proportion of boys that will qualify to run in this event in the state is  area under the normal curve for

substituting values

So

        P(Z<=-2.06) = 0.0197

one that is bigger than 3 1/2

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.