The cutoff time for the fastest 5% of athletes in the men’s group is the same as the 5th percentile. In order to find the time for the 5th percentile we must first find the Z-Score associated with the 5th percentile, which is -1.65. We can then use the Z-Score equation to compute the cutoff time for the 5th percentile as: -1.65 = X – 4313 / 583. When solving for X, we find that X = 3351.05. So therefore, the cutoff time for being in the fastest 5% of athletes in the men’s group is 3351.05 seconds.
The cutoff time for the slowest 10% of athletes in the women’s group is the same as the 90th percentile. In order to find the 90th percentile we must first find the Z-Score associated with the 90th percentile, which is 1.28. We can use the Z-Score equation to compute the cutoff time for the 90th percentile as: 1.28 = X – 5261 / 807. When solving for X, we find that X = 6293.96. Therefore, the cutoff time for being the slowest 10% of athletes in the women’s group is 6293.96 seconds.
To find the percent of passenger vehicles that travel slower than 80mph, we must first find the Z-score. We can find the Z-Score using the equation: Z = 80 – 72.6 / 4.78. This gives us a Z-Score of 1.55, which brings us to about .94 or the 94th percentile. Therefore, about 94% of passenger vehicles travel slower than 80mph on I-5.
In order to find the percentage of vehicles that drive between 60 and 80mph we must first find the percentiles that 60 and 80mph fall at. Again, we can use the Z-Score equation on each vehicle to do this: Z = 60 – 72.6 / 4.78 =-2.6. A Z-Score of -2.63 is associated with the .4th percentile. From Part A, we know that the Z-Score for 80mph is 1.55 and computed that it is associated with the 94th percentile. Therefore, the percentage of vehicles that travel between 60 and 80mph o I-5 is about 94.4%.
Using the Z-Score equation we can find how fast the fastest 5% of passenger vehicles travel. The fastest 5% is the same as the 95th percentile, which has a Z-Score of 1.65. Therefore, 1.65 = X – 72.6 / 4.78 = 80.47. So the fastest 5% of vehicles travel over 80.47mph.
Using the Z-Score equation we find: Z = 70 – 72.6 / 4.78 = -.54. Therefore, about 54% of passenger vehicles travel over the speed limit.
We can test if the heights of the female college students follow the 68-95-99.7 rules first by finding the heights that are one SD above and below the mean, which would be 66.1 and 56.94 respectively. Then we can see how much of the data falls between these numbers. Since 17 out of the 25 fall between these numbers and that is exactly 68% the first part of the rule holds up. We can then test the second part by finding the heights that fall 2 SDs above and below the mean, which are 70.68 and 52.36. Since 24 out of 25 fall between these numbers and that is 96% the second part of the rule also holds up. Finally, we find the heights 3 SDs away, which are 47.78 and 75.26. All of the data falls between this so the final part of the rule holds up and I would say the heights approximately follow the rule.
I believe these heights are a normal distribution because the distribution follows the 68-95-99.7 rule and the probability plot shows little deviation from the line. However, I think it is harder to tell with the histogram because it is such a small sample size.
z <- (72.6-70)/4.78
p <- pnorm(z)
p^5## [1] 0.1763389pnorm(z)## [1] 0.7067562R gives us a probability of .176 or 17.6% that 5 cars pass and none of them speed.
b)To find the average number of cars that would have to pass until a car is speeding can be found by taking the same probability from above (before raising to 5) (.706) and divide the 1 by that probability to get the average…
1/p## [1] 1.414915The average number of cars that would have to drive by before one is speeding is about 1.41. We can now find the standard deviation of this mu by square rooting (1-.706)/(.706^2) in R…
sqrt((1-p)/p^2)## [1] 0.7662046The standard deviation of this distribution is .766.
P(X>=1500|N=15000,p=.09) = ???
I ran this in R Markdown using the pbinom( ) function…
pbinom(1500,15000,.09, lower.tail = FALSE)## [1] 1.173433e-05The probability that at least 1500 will agree to respond to the survey is 1.173e-05 or .000017%.
To find the probability that the first question she gets right is the 3rd question we must take the probability of the first two being wrong times the probability of the 3rd one being right: .75^2 x .25 = .1406. The probability that the first question she gets rights is the third one is .14 or 14%.
To find the probability she gets exactly 3 or 4 right we can use the sum(dbinom)( ) function in R:
sum(dbinom(3:4,5,0.25))## [1] 0.1025391The probability she gets exactly 3 or 4 right is .10 or 10%.
pbinom(3,5,0.25, lower.tail=FALSE)## [1] 0.015625The probability she gets the majority of the questions right is .015 or 1.5%.
choose(9,2)*0.15^3*0.85^7## [1] 0.03895012The probability of her making her 3rd success on her 10th attempt would be .038 or 3.8%.
Even if she made 2 in the first 9 attempts the probability of success on the 10th serve would remain the same, because they are indepndent. So it would still be 15%.
Part a is a binomial distribution in which we are looking for a specific number success (kth,3rd) on a specific number trial (nth, 10th), whereas in part b we are not concerned with the first 9 attempts, just the probability of a success on the 10th.