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. the finishing time for kathy was 272 minutes. what proportion of women who ran the marathon had a finishing time less than kathy's? show your work.

The part of women who took part in the marathon and possessed a finishing time lower than Kathy post rounding it off to the next whole number would be:

- 66%

Given that,

The mean of women's finishing time = 259 minutes

The Standard deviation of women's finishing time = 32 minutes

Kathy's finishing time = 272 minutes

To find,

The part of women who took part in the marathon and possessed a finishing time lower than Kathy post rounding it off to the next whole number = ?

We will first determine Kathy's finishing time through z-score,

z-score

By employing the z-table  

Post rounding off to next hundredth becomes %

Thus, 66% is the correct answer.

Learn more about 'Percentage' here:

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The proportion of women who ran the marathon and had a finishing time less than Kathy's is 66% (rounding to the next whole)

Explanation:

1. Let's review the information given to us to answer the question properly:

μ of finishing time for women = 259 minutes

σ of finishing time for women = 32 minutes

Finishing time for Kathy = 272 minutes

2. What proportion of women who ran the marathon had a finishing time less than Kathy's? Show your work.

Let's calculate the z-score for the finishing time for Kathy, as follows:

z-score = (272 - 259)/32

z-score = 13/32 = 0.41 (rounding to the next hundredth)

Using the z-table, now we can calculate the proportion of women who ran the marathon had a finishing time less than Kathy's:

p (z=0.41) = 0.6591

The proportion of women who ran the marathon and had a finishing time less than Kathy's is 66% (rounding to the next whole)

light-year  

the distance that light travels in a vacuum in one year, equal to about 9.46 trillion km (5.88 trillion mi). light-years are used in measuring interstellar and intergalactic distances.  

parsec  

a unit of distance used in astronomy, equal to about 3.25 light years (3.08 × 1016 meters).  

parallax  

the apparent angular displacement of a celestial body due to its being observed from the surface instead of from the center of the earth (diurnal parallax or geocentric parallax) or due to its being observed from the earth instead of from the sun (annual parallax or heliocentric parallax)  

absolute magnitude  

the magnitude (brightness) of a celestial object as it would be seen at a standard distance of 10 parsecs.  

apparent magnitude  

the magnitude of a celestial object as it is actually measured from the earth.  

redshift  

an increase in the wavelength of radiation emitted by a celestial body as a consequence of the doppler effect.  

the great thing about your imagination is that it doesn't have to make sense! if i were you i would be stuck in space, no communication with earth travelling around and around, you can only guess how long its been since your communication broke down. you land on an undiscovered planet, you do a scientific investigation examining the elements and conditions. you come across alien lifeforms.  

best of luck!