Statistics

Draw a Venn diagram for the following data:

U = numbers under 20

A = even numbers

B = Multiples of 3

Solutions:

Calculate \(P(A \cap B)\)

Solutions:

There are 19 numbers in the entire set.

There are 3 numbers inside the intersection.

So \(P(A \cap B) = \frac{3}{19}\)

There are 400 students in the school, 250 girls and 150 boys. Explain how to take a stratified sample of 40 students in the school.

Solution.

We want to pick 25 girls and 15 boys (so that we have the same proportion as the entire population). One method is to assign all the girls a number from 1-250. Then using a random number generator, generate 25 numbers and then pick those 25 girls.

Repeat the same for the boys: assign them a number from 1-150. Then use your random number generator to generate 15 numbers and then pick those 15 boys.

On a randomly chosen day, each of the 32 students in a class recorded the time (t) in minutes to the nearest minute that it took them to get to school. Find the mean and standard deviation from the following data:

\[\sum t = 1414 \text{ and } \sum t^2 = 69378\]

Solutions:

Mean: \(\frac{\sum t}{n} = \frac{1414}{32} = 44.1875\)

Default Deviation: \(\sqrt{\frac{\sum t^2}{n} - \Big( \frac{\sum t}{n} \Big)^2} = \frac{69378}{32} - (44.1875)^2 = 215.52734375\)

A die is tossed 10 times. The outcome of rolling 5 exhibits a binomial distribution: \(X \sim B(10, \frac{1}{6})\). Calculate \(P(X \leq 3)\).

Solution.

This is as simple as calculating P(X = 0), P(X = 1), P(X = 2) and P(X = 3) and summing them all together. Therefore: \(P(X = 0) = \left( \begin{array} {c} 10 \\ 0 \end{array} \right) (\frac{1}{6})^0 (\frac{5}{6})^{10} = 0.1615055829; \quad P(X=1) = \left( \begin{array} {c} 10 \\ 1 \end{array} \right) (\frac{1}{6})^1 (\frac{5}{6})^9 = 0.3230111658; \quad P(X=2) = \left( \begin{array} {c} 10 \\ 2 \end{array} \right) (\frac{1}{6})^2 (\frac{5}{6})^8 = 0.2907100492; \quad P(X=3) = \left( \begin{array} {c} 10 \\ 3 \end{array} \right) (\frac{1}{6})^3 (\frac{5}{6})^7 = 0.155043596\)

Summing all of these together we get: \(P(X \leq 3) = 0.9302721575\)

A coffee shop claims that a quarter of the cakes sent to them are missing cherries on top. To test this claim, the number of cakes without cherries in a random sample of 40 is recorded. Using a 5% significance level, find the critical region for a two-tailed test of the hypothesis that the probability of a missing cherry is 0.26.

Solutions:

This is a two-tailed test, meaning we will need to look at both ends. Start at a random number on both ends.

Lower end:

P(X ≤ 6) = 0.07452108246

P(X ≤ 5) = 0.03207217407

P(X ≤ 4) = 0.01136083855

As this is a two-tailed test we want to be as close to 0.025 as possible:

P(X ≤ 4) < 0.025 < P(X ≤ 5).

Upper end:

P(X ≥ 16) = 1 - P(X ≤ 15) = 0.03703627013

P(X ≥ 17) = 1 - P(X ≤ 16) = 0.0171086868649

Again we want to be as close to 0.025 as possible so:

P(X ≥ 17) < 0.025 < P(X ≥ 16)

Our critical regions are therefore P(X ≤ 4) and P(X ≥ 17).