[Latexpage]
NOTE: there will be a quiz on Wednesday, based on one of the problems in Confidence Intervals listed below.
Please see this posts for notes: and also more notes and sources on the CLT here: (including yet another link to the resources explaining how to use your graphing calculator to compute special probabilities): Also make sure that you have done the homework in this post! (as usual)
• Additional practice on the exponential distribution and the Central Limit Theorem:
MAT2572additionalExponentialCLTproblems
The problems to do are marked with either a colored dot or a colored line in the margin, except you should omit #8.65(a) and 8.66(a). I will post the answers shortly. Please do these by next Monday at the latest.
• Problems on Confidence Intervals: in your textbook,
p. 309 # 5.3.1, 5.3.2, 5.3.3, 5.3.4, 5.3.13, 5.3.19
Note: in some of these, data is given and you will have to compute the sample mean from the data: just take the ordinary average of the data. Here is a way to enter the data and get the sample mean on TI calculator. and here it is for the Casio graphing calculator.
• For practice identifying the various special discrete probability distributions, use these problems. On first time through, just identify the special distribution (binomial, hypergeometric, poisson, negative binomial, geometric, etc.) and its parameters. Then check your answers against what is given in the last few pages. Testing yourself is one of the best learning methods. You can then go back and test yourself a few days later. Try not to memorize answers based on superficial features of the problem, but rather think about the problem and analyze it to see how it fits with its probability distribution. See these notes:
Math2501DistinguishSpecialProbabilityDistributions
• Please make sure that you have saved your scripts for simulations of the distributions in problems 3.3.1 and 3.3.2, based on the script I did in class and which is also included in this post. You do not have to submit them at this time: we will be using them in a few days to do some hypothesis testing (goodness-of-fit).
Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!
6.50) a) P(-0.002 < X < 0.003) = 1/6
b) P(|X| > 0.005) = P(X > 0.005) + P(X < -0.005) = 2/3
6.51) 1/2
6.54)a) 1 – e-0.2
b) “will not have to be reset in at least 180 days” means X>180: P(X>180) = e-1.5
6.55)
a) ≈ 0.6065
b) ≈ 0.5276
6.56) Don’t forget that the unit of time is 5 hours, so 2 hours is 2/5 of a unit: P(X > 2/5) = e-0.8
6.57) ≈ 0.3012
6.62)
a) ≈ 0.9082
b) ≈ 0.7852
c) ≈ 0.1800
d) ≈ 0.6413
6.63)
a) P(Z>1.14) ≈ 0.1271
b) P(Z>-0.36) ≈ 0.6406
c) P(-0.46< Z <-0.09) ≈ 0.1413
d) P(-0.58< Z <1.12) ≈ 0.5876
6.66) Important problem!
(a) about 68%
(b) about 95%
(c) over 99.7%
(d) close to 100%
These important results are collectively known as “the Empirical Rule” when applied to quantities which have a distribution which is close to normal – in which case they should hold approximately: a nearly normally distributed random variable should have:
about 68% probability to fall within one standard deviation of the mean;
about 95% probability to fall within two standard deviations of the mean;
virtual certainty to fall within three standard deviations of the mean.
6.70
a) z=(44.5-37.6)/4.6 = 1.5; P(z>1.5) ≈ 0.5 – 0.4332 ≈ 0.0668
b) z=(35.0-37.6)/4.6 ≈ -0.57; P(z<-0.57) ≈ 0.5 – 0.2157 ≈ 0.2843
c) z1=-1.65, z2=0.52; P(-1.65< z<0.52) ≈ 0.4505 + 0.1985 ≈ 0.649
6.71a) 0.1056
6.72)
First use the given information to find μ = 73.3; then P(X > 58.3) = approximately 0.9332
6.74)
a) No; μ=np is too small
b) Yes
c) No; n(1-p)=2.4, too small
Note that what the two conditions amount to is that n cannot be too small and p cannot be too close to either 0 or 1
6.75)
a) Yes, because np = 7.5 and n(1-p) = 142.5 both are at least 5
b) 0.0078
c) The percent error is computed by taking the difference between the approximation and the actual value, divided by the actual value, and then convert that result to a percent. $\frac{0.0078 – 0.0036}{0.0036}$ ≈ 117%, obviously unacceptable
6.76) estimate the probability is ≈ 0.2128, error ≈ 0.0033 (less than 1%)
6.77) b(1;150,0.05) ≈ p(1;7.5) = 0.0041 from the tables: this is clearly closer than the normal estimate we got in #6.75 [6.74]. Note: David Moore (who has written several very influential textbooks on Statistics) recommends using a higher cutoff for the normal estimate of the binomial, and you can see the reason by comparing this problem and #6.75.
8.60)
a) $\binom{12}{3}$ = 220
b) $\binom{20}{3}$ = 1140
c) $\binom{50}{3}$ = 19600
8.66) Remember the standard deviation to use is the standard error of the mean, $\frac{\sigma}{\sqrt{n}} = 0.7$, so two standard deviations is 1.4. We are finding the probability that $\bar{X}$ will fall outside of two standard deviations:
$\mu + 2\frac{\sigma}{\sqrt{n}} = 126.6$
$\mu – 2\frac{\sigma}{\sqrt{n}} = 129.4$
b) Compute $P(\bar{X}<126.6) + P(\bar{X}>129.4)$ = less than 0.0456
Note when you change to Z you must use the standard error of the mean.
8.68) $P(\bar{X} > 4.5) = P(Z > 1.8) = 0.5 – 0.4641 \approx 0.0359$
(Recall that for the exponential distribution we have μ = θ and σ = θ)
8.70)
a) $P(\bar{X} > 52.9) = P(Z>1.76) \approx 0.0392$
b) $P(50.5 < \bar{X} < 52.3) = P(-1.05 < Z <1.05) \approx 0.7062$
c) $P(\bar{X} < 50.6) = P(Z < -0.94) \approx 0.1736$
