Biological Statistics Help?

I have been stuck on this problem for HOURS. Please help me understand how to do this because I have about 7 more exactly like it! Thanks!

The results of the 1999 National Health Interview Survey released in 2003 showed that among U.S. adults ages 60 and older, 19% had been told by a doctor that they had some form of cancer. If we use this as the percentage for all adults 65 years old and older living in the U.S., what is the probability that among 65 adults chosen at random more than 25% will have been told by their doctor that they have cancer?

1 Answer

  • 9 years ago
    Favorite Answer

    Hi Rachel -

    Sorry you have to do 7 more! But, the plus side is they are pretty easy to solve.

    This is a "binomial distribution" with n=65 and p=.19. Since n is fairly large, you can use a normal distribution to approximate the answer (that is good news!)

    mean = np = 65(.19) = 12.35

    standard deviation = √(npq) = √(.19*.81*65 = 3.1628

    Now use the normal distribution:

    P(X>25%*65) = P[ z > (16.25 - 12.35)/3.1628 ] = P( z > 1.233)

    Now, look this up in a Standard Normal table and you should get:

    P( z > 1.233) = 0.109

    Hope that helped and good luck with the rest of the problems.

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