These are the solutions for Computer Lab 5.
Please note that answers are given to 4 decimal places of accuracy at most.
\(\overline{X} \sim N(12, 1.3768)\). Note here that \(\dfrac{\sigma^2}{n} = \dfrac{34.42}{25} = 1.3768\).
Example R code for these questions is provided below:
pnorm(10, mean = 12, sd = sqrt(1.3768)) # Note that we have to use the standard deviation, not variance here## [1] 0.04414475So \(P(\overline{X} < 10) = 0.0441\).
pnorm(15, mean = 12, sd = sqrt(1.3768)) - pnorm(4, mean = 12, sd = sqrt(1.3768)) ## [1] 0.994717So \(P(4 \leq \overline{X} \leq 15) = 0.9947\).
1 - pnorm(13, mean = 12, sd = sqrt(1.3768)) ## [1] 0.197039So \(P(\overline{X} > 13)= 0.1970\).
\(P(\overline{X} < 10)\) is the probability that the sample mean number of fuel tanks per craft is less than 10 (given that the population mean is 12 and assuming we took a random sample of 25). This probability is equal to approximately \(4.44\%\).
\(P(4 \leq \overline{X} \leq 15)\) is the probability that the sample mean number of fuel tanks per craft is between 4 and 15 (given that the population mean is 12 and assuming we took a random sample of 25). This probability is equal to approximately \(99.47\%\).
\(P(\overline{X} > 13)\) is the probability that the sample mean number of fuel tanks per craft is greater than 13 (given that the population mean is 12 and assuming we took a random sample of 25). This probability is equal to approximately \(19.7\%\).
What we are really asking here is, find the number \(a\) such that \(P(\overline{X} \leq a) = 0.4\). To solve this in R, we have
qnorm(0.4, mean = 12, sd = sqrt(1.3768)) ## [1] 11.70273Thus, the number of fuel tanks per craft must be 11.7 for this condition to be satisfied.
Similar to 1.4 above, we would like to find the number \(b\) such that \(P(\overline{X} > b) = 0.1\). We can rearrange this so that \(P(\overline{X} \leq b) = 0.9\). To solve this in R, we have
qnorm(0.9, mean = 12, sd = sqrt(1.3768)) ## [1] 13.50374Thus, the number of fuel tanks per craft must be 13.5 for this condition to be satisfied.
Similar to 1.5 above, we would like to find the number \(c\) such that \(P(\overline{X} > c) = 0.05\). We can rearrange this so that \(P(\overline{X} \leq c) = 0.95\). To solve this in R, we have
qnorm(0.95, mean = 12, sd = sqrt(1.3768)) ## [1] 13.93002Thus, the number of fuel tanks per craft must be 13.93 for this condition to be satisfied.
We have
# lower
qnorm(0.025, mean = 12, sd = sqrt(1.3768)) ## [1] 9.700235# upper
qnorm(0.975, mean = 12, sd = sqrt(1.3768)) ## [1] 14.29977So our range of values for fuel tanks per craft is (9.7, 14.3).
Unfortunately, because our sample size is less than 30, and because we do not know the distribution of the population, we cannot actually apply the Central Limit Theorem. Therefore we are unable to determine the distribution of \(\overline{X}\) here.
\(\overline{X} \stackrel{\tiny \text{approx.}}\sim N\left(35, 32.5^2 / 60\right)\). Note here that we are given \(\sigma\), not \(\sigma^2\).
It follows that \(\dfrac{\sigma^2}{n} = \dfrac{32.5^2}{60} \approx 17.60417\). We can say that \(\overline{X}\) is approximately normally distributed due to the Central Limit Theorem, the conditions of which are satisfied since our sample is larger than 30.
Example R code for these questions is provided below:
pnorm(42, mean = 35, sd = sqrt(32.5^2 / 60)) ## [1] 0.9523781So \(P(\overline{X} < 42) = 0.9524\).
pnorm(38, mean = 35, sd = sqrt(32.5^2 / 60)) - pnorm(18.5, mean = 35, sd = sqrt(32.5^2 / 60)) ## [1] 0.7626573So \(P(18.5 \leq \overline{X} \leq 38) = 0.7627\).
1 - pnorm(26, mean = 35, sd = sqrt(32.5^2 / 60)) ## [1] 0.9840251So \(P(\overline{X} > 26)= 0.9840\).
pnorm(30, mean = 35, sd = sqrt(32.5^2 / 60)) +
1- pnorm(36, mean = 35, sd = sqrt(32.5^2 / 60)) ## [1] 0.5225017So \(P(\overline{X} < 30) + P(\overline{X} > 36)= 0.5225\).
\(P(\overline{X} < 42)\) is the probability that the sample mean diameter of an asteroid is less than 42 kms (given that the population mean is 35 and assuming we took a random sample of 60). This probability is equal to approximately \(95.24\%\).
\(P(18.5 \leq \overline{X} \leq 38)\) is the probability that the sample mean diameter of an asteroid is between 18.5 kms and 38 kms (given that the population mean is 35 and assuming we took a random sample of 60). This probability is equal to approximately \(76.27\%\).
\(P(\overline{X} > 26)\) is the probability that the sample mean diameter of an asteroid is greater than 26kms (given that the population mean is 35 and assuming we took a random sample of 60). This probability is equal to approximately \(98.40\%\).
\(P(\overline{X} < 30) + P(\overline{X} > 36)\) is the probability that the sample mean diameter of an asteroid is either less than 30 kms, or greater than 36 kms (given that the population mean is 35 and assuming we took a random sample of 60). This probability is equal to approximately \(52.25\%\).
Given our sample, with \(\overline{X} \stackrel{\tiny \text{approx.}}\sim N\left(35, 32.5^2 / 60 \right)\), we can compute \(P(\overline{X} \geq 42)\). In R, we have
1- pnorm(42, mean = 35, sd = sqrt(32.5^2 / 60)) ## [1] 0.04762194As \(P(\overline{X} \geq 42) = 0.0476\), it is probably safe to say the miner is exaggerating, as based on our sample, there is a less than \(5\%\) chance of observing an asteroid with a diameter of 42 kms or greater. (This could potentially be because the asteroids have gotten smaller due to repeated mining over the years, but we will not give the miner the benefit of the doubt).
Unfortunately, we cannot determine the distribution of \(\overline{X}\) for this new asteroid cluster, as we do not know the population distribution, and the sample size is too small (\(15 < 30\)) to apply the Central Limit Theorem.
No solutions required.
These notes have been prepared by Amanda Shaker and Rupert Kuveke. The copyright for the material in these notes resides with the authors named above, with the Department of Mathematical and Physical Sciences and with La Trobe University. Copyright in this work is vested in La Trobe University including all La Trobe University branding and naming. Unless otherwise stated, material within this work is licensed under a Creative Commons Attribution-Non Commercial-Non Derivatives License BY-NC-ND.