3.2) What percent of a standard normal distribution N(\(\mu\)=0,\(\sigma\)=1) is found in each region?
P(Z>-1.13)=P(-Z<1.13) because we are working with a symetric distribution with a mean at 0. Because Z goes from \(-\infty\) to \(infty\) this is the same as P(Z<1.13) So, in R:
pnorm(1.13)## [1] 0.8707619pnorm(.18)## [1] 0.5714237Once again, to do greater than, we switch to a less than a negatives.
pnorm(-8)## [1] 6.220961e-16We need to break this up, as it represents the space between -.5 and .5 under the normal distribution. So it is P(Z<.5)-P(Z<-.5). In R:
pnorm(.5)-pnorm(-.5)## [1] 0.3829249\(Z=\frac{x-\mu}{\sigma}\)
3.4
In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25-29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups:
\(\mu=4313\),\(\sigma^2\)=583$ for men, \(\mu=5261\), \(\sigma^2=807\)
b.) What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
Leo:
(4948-4313)/583## [1] 1.089194Mary:
(5513-5261)/807## [1] 0.3122677Leo did worse than about 86% of all competitors, Mary was middle of the pack, doing only slightly worse than average.
c.) Did Leo or Mary rank better in their respective groups? Explain your reasoning.
Mary did better, see above.
d.) Leo did worse than 86%, and better than 14%
e.) Mary did worse than 62% and better than 38%
f.) If the distributions were not normal, this could vastly change the above answers. If they were exponential, then being that far behind the mean could be much more significant.
3.18
hts<-c(54,55,56,56,57,58,58,59,60,60,60,61,61,62,62,63,63,63,64,65,65,67,67,69,73)a.) The mean height is 61.52 inches with a standard deviation of 4.58 inches. Use this information to determine if the heights approximately follow the 68-95-99.7% Rule.
\(\mu -\sigma= 56.94\). $+=66.1 $ So \(\frac{4}{25}+\frac{4}{25}=32\%\) are outside of the first standard deviation.
56.94-4.58## [1] 52.36pnorm(52.36,61.25,4.58)## [1] 0.0261262366.1+4.58## [1] 70.68pnorm(70.68,61.25,4.58)## [1] 0.980250652.36-4.58## [1] 47.78pnorm(47.78,61.25,4.58)## [1] 0.00163551970.68+4.58## [1] 75.26pnorm(75.26,61.25,4.58)## [1] 0.9988894It follows the 68-95-99.7% rule perfectly.
b.) Based on the line chart, this appear to be normal, especially given the above information.
3.22
A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others.
a.) What is the probability that the 10th transistor produced is the first with a defect?
.98^9*.02## [1] 0.016674961.7%
b.) What is the probability that the machine produces no defective transistors in a batch of 100?
.98^100## [1] 0.132619613.2%
c.) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?
For exponential distributions,
\(\mu=\frac{1}{\lambda}\), so 50 transistors. \(\sigma=\mu\) so the standard deviation is 50 as well.
d.) Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?
As above, 20 transistors for both mean and standard deviation.
e.) Increasing the probability decreases both the mean and standard distributions.
3.38
While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 kids.
a.) Use the binomial model to calculate the probability that two of them will be boys.
\({3\choose2}\times.51^{2}\times.49^{1}\)=38%
b.) Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate the same probability from part (a) but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match.
There are 3 versions of two boys and one girl.
Boy Boy Girl = .51x.51x.49 =.127
Boy Girl Boy = .51x.49x.51=.127
Girl Boy Boy = .51x.49x.51=.127
Summed up, this equals .38, or 38%, the same as a.)
c.) If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, briefly describe why the approach from part (b) would be more tedious than the approach from part (a).
The addition rule would require \({8\choose3}=56\) posibilities listed out. It is much easier to do the calculation once.
3.42) A not-so-skilled volleyball player has a 15% chance of making the serve, which involves hitting the ball so it passes over the net on a trajectory such that it will land in the opposing team’s court. Suppose that her serves are independent of each other.
a.) What is the probability that on the 10th try she will make her 3rd successful serve?
The odds of making 2 in 9 shots are \({9\choose2}\times(.15)^2\times(.85)^7\). This is then multiplied by .15.
8.8%
in R:
choose(9,2)*(.15^2)*(.85^2)*.15## [1] 0.08778375b.) Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?
15%
c.) Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be di↵erent. Can you explain the reason for this discrepancy?
Part b assumes that the first two servers have already been made, so there is only one trial to to worry about. Part a asks about all 10 serves.