83/84: normalcdf(0.7463,3.731) .
(iii) Normal approximation to the binomial. The conditions for doing a normal approximationwere established in part (ii). Also, μ (^) X = 500(0.1) = 50 and . P (55
≤ X ≤ 75) = P (0.7454 ≤ z ≤ 3.7268) = 0.2279.
All four of these statement are true. However, only III is a statement of the central limit theorem. The
others are true of sampling distributions in general.
- If X is the count of cars with defective pads, then X has B (20, 0.15).
(a) . On the TI-83/84, the solution is given by
binompdf(20,0.15,3).(b) μ (^) X = np = 20(0.15) = 3, .
- μ = 40,000 miles and .
(a) With n = 160, the sampling distribution of will be approximately normally distributed with
mean equal to 40,000 miles and standard deviation 395.28 miles.
(b) .If the manufacturer is correct, there is only about a 0.6% chance of getting an average this low or
lower. That makes it unlikely to be just a chance occurrence, and we should have some doubts
about the manufacturer’s claim.
If X is the number of times you win, then X has B (1000, 0.474). To come out ahead, you must win
more than half your bets. That is, you are being asked for P (X > 500). Because (1000)(0.474) = 474
and 1000(1 – 0.474) = 526 are both greater than 10, we are justified in using a normal approximation
to the binomial. Furthermore, we find that
Now,That is, you have slightly less than a 5% chance of making money if you play 1000 games of
roulette.Using the TI-83/84, the normal approximation is given by normalcdf(500, 10000,474,15.79)
= 0.0498. The exact binomial solution using the calculator is 1-
binomcdf(1000,0.474,500)=0.0467.- μ = 2 lbs = 32 oz and .
(a) With samples of size 35, the central limit theorem tells us that the sampling distribution of is
approximately normal. The mean is 32 oz and standard deviation is 0.845 oz.
(b) In order for to be in the top 10% of samples, it would have to be at the 90th percentile, which