How fast can male college students run a mile? There’s lots of variation, of course. During World War II, physical training was required for male students in many colleges, as preparation for military service. That provided an opportunity to collect data on physical performance on a large scale. A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.14 minutes and standard deviation 0.7 minute. It's good practice to draw a Normal curve on which this mean and standard deviation are correctly located. To do this, draw an unlabeled Normal curve, locate the points where the curvature changes (this is 1 standard deviation from the mean), then add number labels on the horizontal axis. Use the Empirical Rule to answer the following questions. Required:
a. What range of times covers the middle 99.7% of this distribution?
b. What percentage of the these running times are faster than 7.8 minutes?
c. What percent of these runners run the mile between 5.7 and 9.2 minutes?

a)

b)

Using the normal approximation we can assume that this value 7.8 is approximately 1 deviation above the mean so then the percentage of values above is (100-68)/2 = 16%

c)

So we want to find approximately the % between 2 deviation below the mean and 3 deviation above the mean. For the % below two deviations from the mean we have (100-95)/2= 2.5% and for the % above 3 deviations from the mean we got (100-99.7)/2= 0.15% so then the percentage desired would be (100-2.5-0.15)% = 97.35%

Step-by-step explanation:

Let X the random variable that represent the times for the mile run of a population, and for this case we know the distribution for X is given by:

Where and

We can use the z score to find how many deviation we are from the mean with this formula:

Part a

From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean and we can calculate the range like this:

Part b

We can use the z score and we got:

Using the normal approximation we can assume that this value 7.8 is approximately 1 deviation above the mean so then the percentage of values above is (100-68)/2 = 16%

Part c

We can use the z score formula and we got:

So we want to find approximately the % between 2 deviation below the mean and 3 deviation above the mean. For the % below two deviations from the mean we have (100-95)/2= 2.5% and for the % above 3 deviations from the mean we got (100-99.7)/2= 0.15% so then the percentage desired would be (100-2.5-0.15)% = 97.35%

answer: the answer is e

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)