Let  \(P \sim N\left(-2,2^2\right), Q \sim N\left(3,3^2\right), R \sim N\left(5,6^2\right)\)  and  \(Z \sim N(0,1)\).

Given that \(P, Q\) and \(R\) are independent random variables, \(\operatorname{Pr}(3 P+2 Q-R>25)\) is equal to

1. \(\operatorname{Pr}\left(Z>\dfrac{5 \sqrt{3}}{3}\right)\)
2. \(\operatorname{Pr}(Z>5)\)
3. \(\operatorname{Pr}\left(Z>\dfrac{5 \sqrt{66}}{11}\right)\)
4. \(\operatorname{Pr}\left(Z>\dfrac{30}{7}\right)\)

The production of a brand of weed trimmer involves three stages, Stage 1, Stage 2 and Stage 3, which take \(W_1\) hours, \(W_2\) hours and \(W_3\) hours, respectively. Here \(W_1, W_2\) and \(W_3\) are independent random variables, which may be assumed to be normally distributed. Assume that Stage 2 starts immediately after Stage 1 ends and that Stage 3 starts immediately after Stage 2 ends.

The mean, standard deviation and cost at each stage are shown in the table below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Stage} & \textbf{Time (h) } & \textbf{Mean(h)} &\textbf{Standard } &  \textbf{Cost (\$/h)} \\
& & & \rule[-1ex]{0pt}{0pt} \textbf{deviation (h)}\\
\hline \rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& W_1 & 1.0 & 0.3 & 10 \\
\hline \rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& W_2 & 1.5 & 0.4 & 20 \\
\hline \rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt}& W_3 & 2.0 & 0.5 & 15 \\
\hline
\end{array}

1. Find the mean and the variance of the total time to produce one weed trimmer.   (1 mark)
   

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2. Find the variance of the total cost to produce one weed trimmer.   (2 marks)
   

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3. If a single weed trimmer is produced, find the probability that the time spent at Stage 2 will be less than the time spent at Stage 1.
4. Give your answer correct to two decimal places.
5. Use  \(\operatorname{Pr}(-1<Z<1)=0.68\),  where \(Z\) is the standard normal variable with mean 0 and standard deviation 1.  (2 marks)
   

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An office has two coffee machines that operate independently of each other. The time taken for each machine to produce a cup of coffee is normally distributed with a mean of 30 seconds and a standard deviation of 5 seconds. On a particular morning, a cup is produced from each machine.

The probability that the time taken by each coffee machine to produce one cup of coffee will differ by less than 3 seconds is closest to

1. 0.164
2. 0.236
3. 0.329
4. 0.451
5. 0.671

`X` and `Y` are independent random variables where each has a mean of 4 and a variance of 9.

If the random variable  `Z = aX + bY`  has a mean of 8 and a variance of 90, possible values of `a` and `b` are

1. `a = 1, \ b = 1`
2. `a = 4, \ b = −2`
3. `a = 3, \ b = −1`
4. `a = 1, \ b = 3`
5. `a = −2, \ b = 4`

Oranges grown on a citrus farm have a mean mass of 204 grams with a standard deviation of 9 grams

Lemons grown on the same farm have a mean mass of 76 grams with a standard deviation of 3 grams.

The masses of the lemons are independent of the masses of the oranges.

The mean mass and standard deviation, in grams, respectively of a set of three of these oranges and two of these lemons are

1. `764, 3 sqrt 29`
2. `636, 12`
3. `764, sqrt 33`
4. `636, 3 sqrt 10`
5. `636, 33`

`X` is a random variable with a mean of 5 and a standard deviation of 4, and `Y` is a random variable with a mean of 3 and a standard deviation of 2.

If `X` and `Y` are independent random variables and  `Z = X-2Y`, then `Z` will have mean `mu` and standard deviation `sigma` given by

1. `mu = -1, sigma = 0`
2. `mu = -1, sigma = 4 sqrt 2`
3. `mu = 2, sigma = 8`
4. `mu = 2, sigma = 4 sqrt 2`
5. `mu = -1, sigma = 2 sqrt 6`

`U` and `V` are independent normally distributed random variables, where `U` has a mean of 5 and a variance of 1, and `V` has a mean of 8 and a variance of 1. The random variable `W` is defined by  `W = 4U-3V`.

In terms of the standard normal variable  `Z, Pr(W > 5)`  is equivalent to

1. `text(Pr)(Z > (9sqrt7)/7)`
2. `text(Pr)(Z < 1.8)`
3. `text(Pr)(Z < (9sqrt7)/7)`
4. `text(Pr)(Z > 0.2)`
5. `text(Pr)(Z > 1.8)`

A farm grows oranges and lemons. The oranges have a mean mass of 200 grams with a standard deviation of 5 grams and the lemons have a mean mass of 70 grams with a standard deviation of 3 grams.

Assuming masses for each type of fruit are normally distributed, what is the probability, correct to four decimal places, that a randomly selected orange will have at least three times the mass of a randomly selected lemon?

1. 0.0062
2. 0.0828
3. 0.1657
4. 0.8343
5. 0.9172

The scores on the Mathematics and Statistics tests, expressed as percentages, in a particular year were both normally distributed. The mean and the standard deviation of the Mathematics test scores were 71 and 10 respectively, while the mean and the standard deviation of the statistics test scores were 75 and 7 respectively.

Assuming the sets of tests scores were independent of each other, the probability, correct to four decimal places, that a randomly chosen Mathematics score is higher than a randomly chosen Statistics score is

1. 0.2877
2. 0.3716
3. 0.4070
4. 0.7123
5. 0.9088

`X` and `Y` are independent random variables. The mean and the variance of `X` are both 2, while the mean and the variance of `Y` are 2 and 4 respectively.

Given that `a` and `b` are integers, find the values of `a` and `b` if the mean and the variance of  `aX + bY`  are 10 and 44 respectively.  (4 marks)