Question

a gas supplier maintains a team of engineers who are available to deal with leaks reported by customers. most reported leaks can be dealt with quickly but some require a long time. the time (excluding travelling time) taken to deal with reported leaks is found to have a mean of 65 minutes and a standard deviation of 60 minutes. assuming that the times may be modelled by a normal distribution, estimate the probability that: (i) it will take more than 185 minutes to deal with a reported leak; (ii) it will take between 50 minutes and 125 minutes to deal with a reported leak; z - score=—————— percentage/probability =—————— the board of examiners have decided that 80% of all candidates sitting a level maths will obtain a pass grade. the actual exam marks are found to be normally distributed with a mean of 45 and a standard deviation of 7. (a) what is the lowest score a student can get on the exam to be awarded a pass grade? (b) given that 10% of students will achieve an a*, calculate the lowest mark required to get the highest grade. give your answers to the nearest mark. a) z - score for 80% below = —————— actual score for 80% =—————— b) z - score for 10% above = —————— actual score for 10% above=——————

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Solve part (i) of the first problem

Given $\mu = 65$, $\sigma=60$ and $x = 185$. Calculate the z - score:
$z=\frac{185 - 65}{60}=\frac{120}{60}=2$
We want $P(X>185)$, which is equivalent to $P(Z > 2)$. Since the total area under the normal curve is 1, and $P(Z\leq2)$ from the standard normal table is 0.9772, then $P(Z > 2)=1 - 0.9772 = 0.0228$.

Step3: Solve part (ii) of the first problem

For $x_1 = 50$ and $x_2=125$, $\mu = 65$ and $\sigma = 60$.
$z_1=\frac{50 - 65}{60}=\frac{- 15}{60}=-0.25$
$z_2=\frac{125 - 65}{60}=\frac{60}{60}=1$
$P(50$P(-0.25From the standard - normal table, $P(Z < 1)=0.8413$ and $P(Z<-0.25)=0.4013$
So $P(-0.25

Step4: Solve part (a) of the second problem

We want to find the z - score $z$ such that $P(ZUsing the z - score formula $z=\frac{x-\mu}{\sigma}$, with $\mu = 45$, $\sigma = 7$ and $z = 0.84$.
$0.84=\frac{x - 45}{7}$
$x-45=0.84\times7$
$x=45 + 0.84\times7=45+5.88\approx51$

Step5: Solve part (b) of the second problem

We want to find the z - score $z$ such that $P(Z>z)=0.1$, so $P(Z\leq z)=0.9$. Looking up in the standard - normal table, the z - score corresponding to a cumulative probability of 0.9 is approximately $z = 1.28$.
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, with $\mu = 45$, $\sigma = 7$ and $z = 1.28$.
$1.28=\frac{x - 45}{7}$
$x-45=1.28\times7$
$x=45+8.96\approx54$

Answer:

(i) Probability: 0.0228
(ii) Probability: 0.44
(a) Lowest pass - score: 51
(b) Lowest mark for $A^*$: 54