On the TI-83/84, the solution is given by 1–binomcdf(48,0.025,1).
We know that the sampling distribution of will be similar to the shape of the original population for
small n and approximately normal for large n (that’s the central limit theorem). Hence,
(a) if n = 3, the sampling distribution would probably be somewhat skewed to the left.
(b) if n = 30, the sampling distribution should be approximately normal.
Remember that using n ≥ 30 as a rule of thumb for deciding whether to assume normality is for a
sampling distribution just that: a rule of thumb. This is probably a bit conservative. Unless the
original population differs markedly from mound shaped and symmetric, we would expect to see
the sampling distribution of be approximately normal for considerably smaller values of n .
- Since the binomcdf function can’t be used due to calculator overflow, we will use a normal
approximation to the binomial. Let X = the count of heads. Then μ (^) X =(1,000,000)(0.5) = 500,000
(assuming a fair coin) and . Certainly both np and n (1 – p ) are
greater than 10, so the conditions needed to use a normal approximation are present. If we are to have
at least 1000 more heads than tails, then there must be at least 500,500 heads (and, of course no more
than 499,500 tails). Thus, P (there are at least 1000 more heads than tails) = P (X ) ≥ 500500 =
.
The average wait for the first success to occur in a geometric setting is 1/p , where p is the
probability of success on any one trial. In this case, the probability of a girl for any one birth is p =
0.5. Hence, the average wait for the first girl is . So, we have one boy and one girl, on
average, for each two children. The proportion of girls in the population would not change.
- If X is the count of scholarship athletes that graduate from any sample of 55 players, then X has B
(55, 0.20). μ (^) X = 55(0.20) = 11 and .
Putting the numbers 2, 4, 5, and 7 into a list in a calculator and doing 1-Var Stats, we find μ = 4.5
and σ =1.802775638. The set of all samples of size 2 is {(2,4), (2,5), (2,7), (4,5), (4,7), (5,7)} and
the means of these samples are {3, 3.5, 4.5, 4.5, 5.5, 6}. Putting the means into a list and doing 1-Var
Stats to find μ and σ , we get μ = 4.5 (which agrees with the formula) and σ = 1.040833
(which does not agree with . Since the sample is large
compared with the population (that is, the population isn’t at least 10 times as large as the sample),
we use , which does agree with the computedvalue. (Note, students have not been asked to use this formula on the AP exam.)
There are three different ways to do this problem: exact binomial, using proportions, or using a
normal approximation to the binomial. The last two are essentially the same.
(i) Exact binomial . Let X be the count of persons in the sample that have blood type B. Then X has
B (500, 0.10). Also, 11% of 500 is 55 and 15% of 500 is 75. Hence, P (55 ≤ X ≤ 75) = P (X ≤
- – P (X ≤ 54) = binomcdf(500,0.10,75)– binomcdf(500,0.10,54)=0.2475.
(ii) Proportions. We note that μ (^) X = np = 500(0.1) = 50 and n (1 – p ) = 500(0.9) = 90, so we are
OK to use a normal approximation. Also, μ = p = 0.10 and . P (0.11)
< ≤ 0.15) = P (0.7463 ≤ z ≤ 3.731) = 0.2276. On the TI