| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Distribution |
| Type | Multiple independent components |
| Difficulty | Standard +0.3 This is a straightforward application of the exponential distribution with standard techniques: computing probabilities using the CDF, finding percentiles by inverting the CDF, and recognizing that the number of replacements follows a Poisson distribution. All parts are routine calculations requiring no novel insight, though part (c) requires recognizing the Poisson connection which elevates it slightly above pure recall. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) (Let the random variable \(X\) be the lifetime in years of a hair dryer.) | ||
| \(\lambda = \frac{1}{2} = 0.5\) | B1 | |
| \(P(1.8 \leq X \leq 2.5) = \int_{1.8}^{2.5} 0.5e^{-0.5x} dx = [-e^{-0.5x}]_{1.8}^{2.5} = 0.1201\) | M1 A1 A1 | FT their \(\lambda\) for M1A1; Limits required; cao, 3sf required |
| OR | ||
| \(P(1.8 \leq X \leq 2.5) = F(2.5) - F(1.8) = (1 - e^{-0.5 \times 2.5}) - (1 - e^{-0.5 \times 1.8}) = 0.1201\) | (M1) (A1) (A1) | use of cdf; Must be cdf; cao, 3sf required |
| (b) \(P(X \geq k) = 0.2\) or \(P(X < k) = 0.8\) | M1 | FT their \(\lambda\) for M1A1; Attempt to use of |
| \(1 - e^{-0.5k} = 0.8\) | M1 | |
| \(e^{-0.5k} = 0.2\) | ||
| \(-0.5k = \ln 0.2\) | A1 | cao, oe (eg. 2ln5) |
| \(k = 3.2\) | ||
| OR | ||
| \(P(X < k) = \int_0^k 0.5e^{-0.5x} dx = 0.8\) | (M1) | Or integrating between \(k\) and \(\infty\) and setting = 0.2 |
| \([-e^{-0.5x}]_0^k = 0.8\) | ||
| \((-e^{0.5k}) - (-1) = 0.8\) | ||
| \(0.2 = e^{-0.5k}\) | ||
| \(\ln 0.2 = -0.5k\) | (M1) | |
| \(k = 3.2\) | (A1) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (c) Let the random variable \(Y\) be the number of hair dryers replaced in 5 years. \(\text{Po}(2.5)\) over 5 years | B1 | |
| \(P(Y > 3) = 1 - P(Y \leq 3)\) | M1 | FT their 2.5 for discrete distribution |
| \(P(Y > 3) = 1 - 0.75758 = 0.24242\) | A1 | cao |
| (d) Valid assumption e.g. Jon doesn't think the quality is so poor that he buys a different brand. The hair dryers bought weren't from a faulty batch. The quality of the hair dryers hasn't improved in five years. Manufacturing methods haven't changed. The hair dryers still last on average 2 years. The replacement is a new hair dryer (not a used one). | E1 | |
| Total [11] |
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** (Let the random variable $X$ be the lifetime in years of a hair dryer.) | | |
| $\lambda = \frac{1}{2} = 0.5$ | B1 | |
| $P(1.8 \leq X \leq 2.5) = \int_{1.8}^{2.5} 0.5e^{-0.5x} dx = [-e^{-0.5x}]_{1.8}^{2.5} = 0.1201$ | M1 A1 A1 | FT their $\lambda$ for M1A1; Limits required; cao, 3sf required |
| **OR** | | |
| $P(1.8 \leq X \leq 2.5) = F(2.5) - F(1.8) = (1 - e^{-0.5 \times 2.5}) - (1 - e^{-0.5 \times 1.8}) = 0.1201$ | (M1) (A1) (A1) | use of cdf; Must be cdf; cao, 3sf required |
| **(b)** $P(X \geq k) = 0.2$ or $P(X < k) = 0.8$ | M1 | FT their $\lambda$ for M1A1; Attempt to use of |
| $1 - e^{-0.5k} = 0.8$ | M1 | |
| $e^{-0.5k} = 0.2$ | | |
| $-0.5k = \ln 0.2$ | A1 | cao, oe (eg. 2ln5) |
| $k = 3.2$ | | |
| **OR** | | |
| $P(X < k) = \int_0^k 0.5e^{-0.5x} dx = 0.8$ | (M1) | Or integrating between $k$ and $\infty$ and setting = 0.2 |
| $[-e^{-0.5x}]_0^k = 0.8$ | | |
| $(-e^{0.5k}) - (-1) = 0.8$ | | |
| $0.2 = e^{-0.5k}$ | | |
| $\ln 0.2 = -0.5k$ | (M1) | |
| $k = 3.2$ | (A1) | cao |
# Question 3 (continued)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(c)** Let the random variable $Y$ be the number of hair dryers replaced in 5 years. $\text{Po}(2.5)$ over 5 years | B1 | |
| $P(Y > 3) = 1 - P(Y \leq 3)$ | M1 | FT their 2.5 for discrete distribution |
| $P(Y > 3) = 1 - 0.75758 = 0.24242$ | A1 | cao |
| **(d)** Valid assumption e.g. Jon doesn't think the quality is so poor that he buys a different brand. The hair dryers bought weren't from a faulty batch. The quality of the hair dryers hasn't improved in five years. Manufacturing methods haven't changed. The hair dryers **still** last on average 2 years. The replacement is a new hair dryer (not a used one). | E1 | |
| **Total [11]** | | |It is known that the average lifetime of hair dryers from a certain manufacturer is 2 years. The lifetimes are exponentially distributed.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the lifetime of a randomly selected hair dryer is between 1·8 and 2·5 years. [4]
\item Given that 20% of hair dryers have a lifetime of at least $k$ years, find the value of $k$. [3]
\item Jon buys his first hair dryer from the manufacturer today. He will replace his hair dryer with another from the same manufacturer immediately when it stops working. Find the probability that, in the next 5 years, Jon will have to replace more than 3 hair dryers. [3]
\item State one assumption that you have made in part (c). [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2023 Q3 [11]}}