Question 2 Begin writing on a NEW page a) When a particular variety of radish is grown without fertilizer, the weights of the radishes produced are normally distributed with mean $40 \mathrm{~g}$ and standard deviation $10 \mathrm{~g}$. When these radishes are grown in the same conditions but with fertilizer added, their weights are also normally distributed, with a mean $140 \mathrm{~g}$ and standard deviation $40 \mathrm{~g}$. One radish grown with fertilizer and one radish grown without fertilizer are selected at random. Using diagrams (these must be included in your answer) and the empirical rule, find the probability that both radishes weigh more than $60 \mathrm{~g}$. Give your answer correct to 4 decimal places. b) The random variable $T$ represents the lifetime in years of a component of a solar cell. The probability density function of $T$ is: \[ f(t)=0.4 e^{-0.4 t} \text { where } t \geq 0 \] i) Find the probability that a particular component of the solar cell fails within one year. Give your answer correct to 2 decimal places. ii) Each solar cell has 5 of these components which operate independently of each other. The cell will work provided at least 3 of the components continue to work. Find the probability that a solar cell will still operate after one year. Give your answer correct to 4 decimal places. c) The waiting time (in minutes) at a dental surgery is represented by a random variable $X$ which has a probability density function: \[ f(x)=\left\{\begin{array}{c} 0, \quad \text { if } x<0 \\ \frac{x}{40000}\left(400-x^{2}\right) \quad \text { if } 0 \leq x \leq 20 \\ 0, \quad \text { if } x>20 \end{array}\right. \] The surgery conducted a customer satisfaction survey that involved rating the surgery from 1 star (extremely dissatisfied) to 5 stars (extremely satisfied). The results of the survey were normally distributed, with a mean of 3 stars and a standard deviation of 0.5 stars. $8 \%$ of patients with a waiting time greater than 18 minutes gave a rating of 4 or more stars. Use the empirical rule when required to answer the following questions: i) Determine the most common waiting time. Give your answer to 2 decimal places. ii) Find $P(X>18)$, Give your answer as an exact fraction. iii) Hence, find the probability that a patient waited for more than 18 minutes given that they gave 4 or more stars on the survey. Give your answer to 4 decimal places. Marks 2 1 2 2 1 5
Solution
Step 1 ：First, we need to find the probability that a radish grown without fertilizer weighs more than 60g. Using the empirical rule, we know that 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
Step 2 ：For radishes grown without fertilizer, the mean is 40g and the standard deviation is 10g. So, one standard deviation above the mean is 50g, two standard deviations above the mean is 60g, and three standard deviations above the mean is 70g.
Step 3 ：Since 60g is two standard deviations above the mean, we know that 95% of the data falls within this range. Therefore, the probability that a radish grown without fertilizer weighs more than 60g is 1 - 0.95/2 = 0.025.
Step 4 ：Next, we need to find the probability that a radish grown with fertilizer weighs more than 60g. For radishes grown with fertilizer, the mean is 140g and the standard deviation is 40g. So, one standard deviation below the mean is 100g, two standard deviations below the mean is 60g, and three standard deviations below the mean is 20g.
Step 5 ：Since 60g is two standard deviations below the mean, we know that 95% of the data falls within this range. Therefore, the probability that a radish grown with fertilizer weighs more than 60g is 1 - 0.95/2 = 0.975.
Step 6 ：Finally, since the two radishes are selected independently, we can multiply the probabilities to find the probability that both radishes weigh more than 60g: 0.025 * 0.975 = 0.024375.
Step 7 ：\(\boxed{0.0244}\)
