According to the NIH, 32% of all women will fracture their hip by age 90. If 8 women age 90 are selected at random, what is the probability that exactly 5 of them will have suffered a hip fracture?
- Step 1: begin by turning 32% into .32, This will be your success which is our (P).
- Step 2: next we need to find the failure number , so you have to subtract 1 – .32 = .68 this will be our (q)
- Step 3: Since 8 women are selected at random, this value will be our (n) or sample group
- Step 4: Since the question is asking exactly 5 out of the 8 will have suffered a hip fracture, 5 will be our (x)
- Step5: now that we have all our values we have to check to see if n*p and n*q are greater than (>) 5 (this is a very important step and it must always be done for this type of problem to insure that you end up using the correct formula). So it will look like this np= 8 * .32= 2.56 nq=8 * .68= 5.44 **NOTE** since one of our values is not greater than 5 (np=2.56) we have to use the binomial distribution equation: nCx * p^x * q^n-x
- Step 6: Now plug these values into the equation like so: n=8 , p= .32, q=.68, x=5 8C5 ( .32 )^5 ( .68 ) ^8-5 = 56 ( .32 )^5 ( .68 ) ^3 = .059
Hope my explanation was helpful and good luck on the final EVERYONE !!!! 🙂