I am in an AP Statistics Class and am having trouble on a question and would like help if anyone can. Thanks?

3 Comments

1. Nigel M says:

   May 14, 2010 at 7:37 pm

   There is not sufficient information to answer the question. What is missing is some sort of confidence interval for the 80% effectiveness number: ie 80% with 1s of 5%.
   http://en.wikipedia.org/wiki/Normal_distribution
2. piper411 says:

   May 14, 2010 at 7:48 pm

   I’m in AP Stat, too, so I’m assuming we’re doing the same unit right now. This is a binomial distribution b/c there are a fixed number of trials (100) and you are trying to find P(75 P(X=x)=(n/x)*p^x*(1-p)^n-x
   n/x= nCx (combination, to figure out how many possible orders you can have x successes in n trials)
   n=number of trials
   p=probability
   x= number of successes in n trials
   So, for 75 successes, P(X=75)=(100/75)*.80^75*(1-.80)^100-75= .04387
   Keep doing this for all numbers from 75 to 90 and add up your answers. This will work. It’s probably not the fastest way, but it’s the only way I know. I hope this looks familiar to you, because otherwise you are going to be lost.
3. Merlyn says:

   May 14, 2010 at 8:11 pm

   Let X be the number of patients obtaining some relief. X has the binomial distribution with n = 100 trials and success probability p = 0.8 .

   In general, if X has the binomial distribution with n trials and a success probability of p then
   P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
   for values of x = 0, 1, 2, …, n
   P[X = x] = 0 for any other value of x.

   The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n – x failures.
   Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.

   X ~ Binomial( n , p )

   the mean of the binomial distribution is n * p = 80
   the variance of the binomial distribution is n * p * (1 – p) = 16
   the standard deviation is the square root of the variance = √ ( n * p * (1 – p)) = 4

   To solve this question would require solving a large sum. It is easy to do with software, not so much fun to do by hand. However, we can approximate the probability by using the Normal approximation to the binomial.

   To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.

   In this case you have:
   n * p = 100 * 0.8 = 80 expected success
   n * (1 – p) = 100 * 0.2 = 20 expected failures

   We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.

   If X ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ

   X ~ Binomial(n = 100 , p = 0.8 )
   Xn ~ Normal( μ = 80 , σ² = 16 )
   Xn ~ Normal( μ = 80 , σ = 4 )

   I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.

   The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.

   P( X < x) ≈ P( Xn < (x - 0.5) )
   P( X > x) ≈ P( Xn > (x + 0.5) )
   P( X ≤ x) ≈ P( Xn ≤ (x + 0.5) )
   P( X ≥ x) ≈ P( Xn ≥ (x – 0.5) )
   P( X = x) ≈ P( (x – 0.5) < Xn < (x + 0.5) )
   P( a ≤ X ≤ b ) ≈ P( (a – 0.5) < Xn < (b + 0.5) )
   P( a ≤ X < b ) ≈ P( (a – 0.5) < Xn < (b – 0.5) )
   P( a < X ≤ b ) ≈ P( (a + 0.5) < Xn < (b + 0.5) )
   P( a < X < b ) ≈ P( (a + 0.5) < Xn < (b – 0.5) )

   In the work that follows X has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.

   Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn – μ ) / σ

   INCLUDING BOTH END POINTS
   P( 75 ≤ X ≤ 90 ) =

   90
   ∑ P(X = x) = 0.910191
   x = 75

   ≈ P( 74.5 < Xn < 90.5 )
   = P( ( 74.5 – 80 ) / 4 < Z < ( 90.5 – 80 ) / 4 )
   = P( -1.375 < Z < 2.625 )
   = P( Z < 2.625 ) – P( Z < -1.375 )
   = 0.9956676 – 0.08456572
   = 0.9111018

   INCLUDE LOWER END POINT, EXCLUDE UPPER END POINT
   P( 75 ≤ X < 90 ) = 0.9068282

   89
   ∑ P(X = x) = 0.9068282
   x = 75

   ≈ P( 74.5 < Xn < 89.5 )
   = P( ( 74.5 – 80 ) / 4 < Z < ( 89.5 – 80 ) / 4 )
   = P( -1.375 < Z < 2.375 )
   = P( Z < 2.375 ) – P( Z < -1.375 )
   = 0.9912255 – 0.08456572
   = 0.9066598

   EXCLUDE LEFT END POINT, INCLUDE RIGHT END POINT
   P( 75 < X ≤ 90 ) = 0.8663132

   90
   ∑ P(X = x) = 0.8663132
   x = 76

   ≈ P( 75.5 < Xn < 90.5 )
   = P( ( 75.5 – 80 ) / 4 < Z < ( 90.5 – 80 ) / 4 )
   = P( -1.125 < Z < 2.625 )
   = P( Z < 2.625 ) – P( Z < -1.125 )
   = 0.9956676 – 0.1302945
   = 0.865373

   PURE INEQUALITY
   P( 75 < X < 90 ) = 0.8629504

   89
   ∑ P(X = x) = 0.8629504
   x = 76

   ≈ P( 75.5 < Xn < 89.5 )
   = P( ( 75.5 – 80 ) / 4 < Z < ( 89.5 – 80 ) / 4 )
   = P( -1.125 < Z < 2.375 )
   = P( Z < 2.375 ) – P( Z < -1.125 )
   = 0.9912255 – 0.1302945
   = 0.860931