SPS SPS FM Statistics (SPS FM Statistics) 2025 April

It is known that, under standard conditions, 12% of light bulbs from a certain manufacturer have a defect. A quality improvement process has been implemented, and a random sample of 200 light bulbs produced after the improvements was selected. It was found that 15 of the 200 light bulbs were defective.

1. State one assumption needed in order to use a binomial model for the number of defective light bulbs in the sample. [1]
2. Test, at the 5% significance level, whether the proportion of defective light bulbs has decreased under the new process. [7]

In a study of reaction times, 25 participants completed a test where their reaction times (in milliseconds) were recorded. The results are shown in the stem-and-leaf diagram below: 20 | 3 5 7 9 21 | 0 2 5 6 8 22 | 1 3 4 5 7 9 23 | 0 2 5 8 24 | 1 4 6 7 25 | 2 5 Key: 21 | 0 represents a reaction time of 210 milliseconds

1. State the median reaction time. [1]
2. Calculate the interquartile range of these reaction times. [2]
3. Find the mean and standard deviation of these reaction times. [3]
4. State one advantage of using a stem-and-leaf diagram to display this data rather than a frequency table. [1]
5. One participant completed the test again and recorded a reaction time of 195 milliseconds. Add this result to the stem-and-leaf diagram and state the effect this would have on:
   1. the median
   2. the mean
   3. the standard deviation
   [4]
6. Explain why the interquartile range might be preferred to the standard deviation as a measure of spread in this context [2]

Miguel has six numbered tiles, labelled 2, 2, 3, 3, 4, 4. He selects two tiles at random, without replacement. The variable \(M\) denotes the sum of the numbers on the two tiles.

1. Show that \(P(M = 6) = \frac{1}{3}\) [2]

The table shows the probability distribution of \(M\)

\(m\)	4	5	6	7	8
\(P(M = m)\)	\(\frac{1}{15}\)	\(\frac{4}{15}\)	\(\frac{1}{3}\)	\(\frac{4}{15}\)	\(\frac{1}{15}\)

Miguel returns the two tiles to the collection. Now Sofia selects two tiles at random from the six tiles, without replacement. The variable \(S\) denotes the sum of the numbers on the two tiles that Sofia selects.

1. Find \(P(M = S)\) [3]
2. Find \(P(S = 7 | M = S)\) [4]

The discrete random variable \(X\) has a geometric distribution. It is given that \(\text{Var}(X) = 20\). Determine \(P(X \geq 7)\). [6]

An examination paper consists of 8 questions, of which one is on geometric distributions and one is on binomial distributions.

1. If the 8 questions are arranged in a random order, find the probability that the question on geometric distributions is next to the question on binomial distributions. [2]

Four of the questions, including the one on geometric distributions, are worth 7 marks each, and the remaining four questions, including the one on binomial distributions, are worth 9 marks each. The 7-mark questions are the first four questions on the paper, but are arranged in random order. The 9-mark questions are the last four questions, but are arranged in random order. Find the probability that

1. the questions on geometric distributions and on binomial distributions are next to one another, [2]
2. the questions on geometric distributions and on binomial distributions are separated by at least 2 other questions. [3]

The random variable \(X\) represents the weight in kg of a randomly selected male dog of a particular breed. \(X\) is Normally distributed with mean 30.7 and standard deviation 3.5.

1. Find the 90th percentile for the weights of these dogs. [2]
2. Five of these dogs are chosen at random. Find the probability that exactly four of them weighs at least 30 kg. [3]

The weights of females of the same breed of dog are Normally distributed with mean 26.8 kg.

1. Given that 5% of female dogs of this breed weigh more than 30 kg, find the standard deviation of their weights. [3]
2. Sketch the distributions of the weights of male and female dogs of this breed on a single diagram. [3]

The random variable \(y\) has probability density function f(y) given by $$f(y) = \begin{cases} ky(a - y) & 0 \leq y \leq 3 \\ 0 & \text{otherwise} \end{cases}$$ where \(k\) and \(a\) are positive constants.

1. 1. Explain why \(a \geq 3\) [1]
   2. Show that \(k = \frac{2}{9(a - 2)}\) [3]

Given that \(E(Y) = 1.75\)

1. Find the values of a and k. [4]
2. Write down the mode of Y [1]