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Probability, MET2 2024 VCAA 19 MC

Consider the normal random variable \(X\) that satisfies  \( \text{Pr} (X < 10) = 0.2\)  and  \(\text{Pr}(X > 18) = 0.2 \).

The value of  \(\text{Pr}(X<12)\)  is closest to

  1. 0.134
  2. 0.297
  3. 0.337
  4. 0.365
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Pr}\left(z<\dfrac{10-\mu}{\sigma}\right)=0.2\)

\(\text{Pr}\left(z\leq\dfrac{18-\mu}{\sigma}\right)=0.8\)

\(\text{Solve simultaneous equations (by CAS):}\)

\(z_1=\dfrac{10-\mu}{\sigma}\ , z_2=\dfrac{18-\mu}{\sigma}\)

\(\therefore\ \text{Pr}(X<12)\approx 0.336947\)

\(\Rightarrow C\)

Mean mark 57%.

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.

  1. Find  \(\Pr(D>6.8)\), correct to four decimal places.   (1 mark)
  2. Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
  3. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.

  1. Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
  4. Give your answer correct to four decimal places.   (2 marks)

A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.

  1. Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
  2. Give your answer correct to four decimal places.   (2 marks)
  3. The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
  4. Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places.   (2 marks)
  5. An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
  6. The confidence interval is (0.7382, 0.9493), correct to four decimal places.
  7. Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer.   (2 marks)

A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
 

\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)         &\ \ 30 \leq v \leq 3\pi^2+30 \\
0         &\ \ \text{elsewhere}
\end{cases}\)
 

  1. Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. Find the exact mean serving speed for grade A balls, in metres per second.   (1 mark)

The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).

A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).

  1. If the mean serving speed for a grade B ball is \(2\pi^2+8\) metres per second, find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(0.1587\)

b.    \(d\approx6.83\)

c.    \(0.9938\)

d.    \(0.9998\)

e.    \(0,8960\)

f.    \(0.06\)

g.    \(90\%\)

h.    \(0.1345\)

i.    \(3(\pi^2+4)\)

j.    \(a=\dfrac{3}{2}, b=\dfrac{2}{3}\)

Show Worked Solution

a.    \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(6.8, \infty, 6.7, 0.1)]\)

\(\Pr(D>6.8)=0.15865…\approx0.1587\)
 

b.    \(\Pr(D<d)=0.90\)  

\(\Pr\Bigg(Z<\dfrac{d-6.7}{0.1}\Bigg)=0.90\)

\(\text{Using CAS: }[\text{invNorm}(0.9, 6.7, 0.1)]\)

\(\therefore\ d=6.828155…\approx6.83\ \text{cm}\)

 

c.   \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(-\infty, 6.95, 6.7, 0.1)]\)

\(\Pr(D<6.95)=0.99379…\approx0.9938\)
 

d.    \(\text{Binomial:}\to  n=4, p=0.99379…\)

\(X=\text{number of balls}\)

\(X\sim \text{Bi}(4, 0.999379 …)\)

\(\text{Using CAS: }[\text{binomCdf}(4, 0.999379 …, 3, 4)]\)

\(\Pr(X\geq 3)=0.99977 …\approx 0.9998\)

 

e.    \(\Pr(\text{Grade A|Fits})\) \(=\Pr(6.54<D<6.86|D<6.95)\)
    \(=\dfrac{\Pr(6.54<D<6.86)}{\Pr(D<6.95)}\)
  \(\text{Using CAS: }\) \(=\Bigg[\dfrac{\text{normCdf}(6.54, 6.86, 6.7, 0.1)}{\text{normCdf}(-\infty, 6.95, 6.7, 0.1)}\Bigg]\)
    \(=\dfrac{0.89040…}{0.99977…}\)
    \(=0.895965…\approx 0.8960\)

  

f.    \(\text{Normally distributed → symmetrical}\)

\(\text{Pr ball diameter outside the 99% interval}=1-0.99=0.01\)

\(D\sim N(6.7, \mu^2)\)

\(\Pr(6.54<D<6.86)>0.99\)

\(\therefore\ \Pr(D<6.54)<\dfrac{1-0.99}{2}\)

\(\text{Find z score using CAS: }\Bigg[\text{invNorm}\Bigg(\dfrac{1-0.99}{2},0,1\Bigg)\Bigg]=-2.575829…\)

\(\text{Then solve:} \ \dfrac{6.54-6.7}{\sigma}<-2.575829…\)

\(\therefore\ \sigma<0.0621\approx 0.06\)


♦♦ Mean mark (f) 35%.
MARKER’S COMMENT: Incorrect responses included using \(\Pr(D<6.86)=0.99\). Many answers were accepted in the range \(0<\sigma\leq0.06\) as max value of SD was not asked for provided sufficient working was shown.

g.   \(\hat{p}=\dfrac{0.7382+0.9493}{2}=0.84375\)

\(\text{Solve the following simultaneous equations for }z, \hat{p}=0.84375\)

\(0.84375-z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.7382\)      \((1)\)
\(0.84375+z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.9493\) \((2)\)
\(\text{Equation}(2)-(1)\)    
\(2z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.2111\)  
\(z=\dfrac{0.10555}{\sqrt{\dfrac{0.84375(1-0.84375)}{32}}}\) \(=1.64443352\)  

 
\(\text{Alternatively using CAS: Solve the following simultaneous equations for }z, \hat{p}\)

\(\hat{p}-z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.7382\)
\(\text{and}\)  
\(\hat{p}+z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.9493\)

\(\rightarrow \ z=1.64444…\ \text{and}\ \hat{p}=0.84375\)
 

\(\therefore\ \text{Using CAS: normCdf}(-1.64443352, 1.64443352, 0 , 1)\)

\(\text{Level of Confidence} =0.89999133…=90\%\)


♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: \(\hat{p}\) was often calculated incorrectly with \(\hat{p}=0.8904\) frequently seen.

h.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{50}^{3\pi^2+30} \frac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=0.1345163712\approx 0.1345\)

i.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{30}^{3\pi^2+30} v.\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=3(\pi^2+4)=3\pi^2+12\)

j.  \(\text{When the function is dilated in both directions, }a\times b=1\)
 

\(\text{Method 1 : Simultaneous equations}\)

\(g(w) = \begin {cases}
\dfrac{b}{6\pi}\sin\Bigg(\sqrt{\dfrac{\dfrac{w}{b}-30}{3}}\Bigg)         &\ \ 30b \leq v \leq b(3\pi^2+30) \\
\\ 0         &\ \ \text{elsewhere}
\end{cases}\)

\(\text{Using CAS: Define }g(w)\ \text{and solve the simultaneous equations below for }a, b.\)
 

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} g(w)\,dw=1\)

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} w.g(w)\,dw=2\pi^2+8\)

\(\therefore\ b=\dfrac{2}{3}\ \text{and}\ a=\dfrac{3}{2}\)

 
\(\text{Method 2 : Transform the mean}\)

 

\(\text{Area}\) \(=1\)
\(\therefore\ a\) \(=\dfrac{1}{b}\)
\(\to\ b\) \(=\dfrac{E(W)}{E(V)}\)
  \(=\dfrac{2\pi^2+8}{3\pi^2+12}\)
  \(=\dfrac{2(\pi^2+4)}{3(\pi^2+4)}\)
  \(=\dfrac{2}{3}\)
 
\(\therefore\ a\) \(=\dfrac{3}{2}\)

♦♦♦ Mean mark (j) 10%.

Statistics, MET2 2020 VCAA 3

A transport company has detailed records of all its deliveries. The number of minutes a delivery is made before or after its schedule delivery time can be modelled as a normally distributed random variable, `T`, with a mean of zero and a standard deviation of four minutes. A graph of the probability distribution of `T` is shown below.
 

  1. If  `"Pr"(T <= a)=0.6`, find `a` to the nearest minute.   (1 mark)

  2. Find the probability, correct to three decimal places, of a delivery being no later than three minutes after its scheduled delivery time, given that it arrives after its scheduled delivery time.   (2 marks)

  3. Using the model described above, the transport company can make 46.48% of its deliveries over the interval  `-3 <= t <= 2`.
  4. It has an improved delivery model with a mean of `k` and a standard deviation of four minutes.
  5. Find the values of `k`, correct to one decimal place, so that 46.48% of the transport company's deliveries can be made over the interval  `-4.5 <= t <= 0.5`   (3 marks)

A rival transport company claims that there is a 0.85 probability that each delivery it makes will arrive on time or earlier.

Assume that whether each delivery is on time or earlier is independent of other deliveries.

  1. Assuming that the rival company's claim is true, find the probability that on a day in which the rival company makes eight deliveries, fewer than half of them arrive on time or earlier. Give your answer correct to three decimal places.   (2 marks)

  2. Assuming that the rival company's claim is true, consider a day in which it makes `n` deliveries.
    1. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier.   (1 mark)

    2. Hence, or otherwise, find the minimum value of `n` such that there is at least a 0.95 probability that one or more deliveries will not arrive on time or earlier.   (1 mark)
  3. An analyst from a government department believes the rival transport company's claim is only true for deliveries made before 4 pm. For deliveries made after 4 pm, the analyst believes the probability of a delivery arriving on time or earlier is `x`, where  `0.3 <=x <= 0.7`
  4. After observing a large number of the rival transport company's deliveries, the analyst believes that the overall probability that a delivery arrives on time or earlier is actually 0.75
  5. Let the probability that a delivery is made after 4 pm be `y`.
  6. Assuming that the analyst's belief are true, find the minimum and maximum values of `y`.   (2 marks)

Show Answers Only

  1. `a= 1\ text(minute)`
  2. `0.547`
  3. `k=-1.5, -2.5`
  4. `0.003`
  5.  i. `1-0.85^n`
  6. ii. `19`
  7. `2/3`

Show Worked Solution

a.   `T\ ~\ N(0, 4^2)`

`text(Solve (by CAS): Pr)(T<=a) = 0.6`

`:. a= 1\ text(minute)`
 

b.    `text{Pr}(T <= 3∣T > 0)` `=(text{Pr}(0 < T <= 3))/(text{Pr}(T > 0))`
    `=(0.27337 dots)/(0.5)`
    `=0.547\ \ text{(to 3 d.p.)}`

 

c.   `text(Given)\ \ text{Pr}(-3 <= T <= 2) = 0.4648`

`sigma = 4 text{minutes}`

`=> \ text{Pr}(-4.5 <= T – 1.5 <= 0.5) = 0.4648`

`=> k=-1.5`

`text(By symmetry of the normal distribution)`

`text{Pr}(-2 <= T <= 3) = text{Pr}(-3 <= T <= 2) = 0.4648`

`=> \ text{Pr}(-4.5 <= T – 2.5 <= 0.5) = 0.4648`

`=> k=-2.5`

`:. k=-1.5, -2.5`

 

d.   `text{Let}\ \ X\ ~\ text{Bi}(8, 0.85)`

`text(Solve (by CAS):)`

`text{Pr}(X<=3) = 0.003\ \ text{(to 3 d.p.)}`
 

e.i.   `text{Pr(at least 1 delivery is late)}`

`= 1-\ text{Pr(all deliveries are on time)}`

`=1-0.85^n`
 

e.ii.   `text{Solve for}\ n:`

`1-0.85^n` `<0.95`  
`n` `>18.43…`  

 
`:.n_min=19`
 

f.   `text{Pr(delivery made after 4pm)} = y`

`=>\ text{Pr(delivery made before 4pm)} = 1-y`

`0.85(1-y)+xy` `=0.75`  
`y` `=-(0.1)/(x-0.85)`  
  `=(2)/(17-20 x)`  

 
`text(Given ) 0.3<=x<=0.7:`

`y_min = (2)/(17-20 xx 0.3) = 2/11`

`y_max = (2)/(17-20 xx 0.7) = 2/3`

Statistics, MET1-NHT 2019 VCAA 6b

Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

Albin observes a total of 12 heads from the 18 tosses.

Based on this sample, find the approximate 90% confidence interval for the probability of observing a head when this coin is tossed. Use the `z` value `33/20`.  (2 marks)

Show Answers Only

`(29/60, 51/60)`

Show Worked Solution

`hat p = 12/18 = 2/3`

`text(90% confidence interval)`

`= (2/3 – 33/20 sqrt((2/3 xx 1/3)/18), 2/3 + 33/20 sqrt((2/3 xx 1/3)/18))`

`= (2/3 – 33/20 sqrt(1/81), 2/3 + 33/20 sqrt(1/81))`

`= (2/3 – 11/60, 2/3 + 11/60)`

`= (29/60, 51/60)`

Probability, MET2 2009 VCAA 3

The Bouncy Ball Company (BBC) makes tennis balls whose diameters are normally distributed with mean 67 mm and standard deviation 1 mm. The tennis balls are packed and sold in cylindrical tins that each hold four balls. A tennis ball fits into such a tin if the diameter of the ball is less than 68.5 mm.

  1. What is the probability, correct to four decimal places, that a randomly selected tennis ball produced by BBC fits into a tin?  (2 marks)

BBC management would like each ball produced to have diameter between 65.6 and 68.4 mm.

  1. What is the probability, correct to four decimal places, that the diameter of a randomly selected tennis ball made by BBC is in this range?  (2 marks)

    1. What is the probability, correct to four decimal places, that the diameter of a tennis ball which fits into a tin is between 65.6 and 68.4 mm?  (1 mark)

    2. A tin of four balls is selected at random. What is the probability, correct to four decimal places, that at least one of these balls has diameter outside the desired range of 65.6 to 68.4 mm?  (2 marks)

BBC management wants engineers to change the manufacturing process so that 99% of all balls produced have diameter between 65.6 and 68.4 mm. The mean is to stay at 67 mm but the standard deviation is to be changed.

  1. What should the new standard deviation be (correct to two decimal places)?  (3 marks)

Show Answers Only

a.  `0.9332`

b.  `0.8385`

c.i.  `0.8985`

c.ii.  `0.3482`

d.  `0.54\ text(mm)`

Show Worked Solution

a.   `text(Let)\ \ X = text(diameter),\ \ X ∼ text(N) (67, 1^2)`

`text(Pr) (X < 68.5) = 0.9332\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (– oo, 68.5, 67, 1)]`

 

b.   `text(Pr) (65.6 < X < 68.4)`

`= 0.8385\ \ text{(to 4 d.p.)}`

`[text(CAS: normCdf) (65.6, 68.4, 67, 1)]`

 

c.i.   `text(Pr) (65.6 < X < 68.4 \|\ X < 68.5)`

♦ Mean mark 42%.

`= (text{Pr} (65.6 < X < 68.4))/(text{Pr} (X < 68.5))`

`= (0.838487…)/(0.933193…)`

`= 0.8985\ \ text{(to 4 d.p.)}`

 

  ii.  `text(Let)\ \ Y = text(Number of balls with diameter outside range)`

♦♦ Mean mark 30%.

`Y ∼ text(Bi)(4, 1 – 0.8985…) -> Y ∼ text(Bi) (4, 0.101486)`

`text(Pr) (Y >= 1) = 0.3482\ \ text{(to 4 d.p.)}`

 

d.   `X = text(diameter),\ \ X ∼ text(N) (67, sigma^2)`

♦♦ Mean mark 35%.

 vcaa-graphs-fur2-2009-3di

`text(Pr) (Z < a)` `= 0.005`
`a` `= – 2.5758`

 

`text(Relate 65.6 to its corresponding)\ \ z text(-score):`

`– 2.5758…` `= (65.6 – 67)/sigma`
`:. sigma` `= 0.54\ text(mm)\ \ text{(to 2 d.p.)}`

Probability, MET2 2011 VCAA 2*

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function
 

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y-4)),y > 4):}`
 

  1. Find correct to four decimal places

    1. `text(Pr)(3 <= X <= 5)`   (1 mark)

    2. `text(Pr)(3 <= Y <= 5)`   (3 marks)

  2. Find the mean of `Y`, correct to three decimal places.   (3 marks)

  3. It can be shown that  `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce.   (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

  1. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.
  2. Find, correct to four decimal places, the probability that it was produced by machine A.   (3 marks)

Show Answers Only

a.i.  `0.4938`

a.ii. `0.4155`

b.    `4.333`

c.    `0.1812`

d.    `0.4103`

Show Worked Solution

a.i.   `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(3 <= Y <= 5)` `= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
    `= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y-4)))dy`
    `= 0.4155\ \ text{(4 d.p.)}`

 

b.    `text(E)(Y)` `= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y-4))dy`
    `= 4.333\ \ text{(3 d.p.)}`

 

c.   `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`
 

`text(Solution 2)`

`text(Pr)(W = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.1812`

♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

 

d.   
`text(Pr)(A | L)` `= (text(Pr)(AL))/(text(Pr)(L))`
  `= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
  `= 0.4103\ \ text{(4 d.p.)}`

Probability, MET2 2010 VCAA 13 MC

The continuous random variable `X` has a normal distribution with mean 20 and standard deviation 6. The continuous random variable `Z` has the standard normal distribution.

The probability that `Z` is between – 2 and 1 is equal to

  1. `text(Pr) (18 < X < 21)`
  2. `text(Pr) (14 < X < 32)`
  3. `text(Pr) (14 < X < 26)`
  4. `text(Pr) (8 < X < 32)`
  5. `text(Pr) (X > 14) + text(Pr) (X < 26)`
Show Answers Only

`B`

Show Worked Solution

`text(When)\ \ Z=-2,`

`X=mu – 2σ = 20-2xx6 = 8`

`text(When)\ \ Z=1,`

`X=mu + σ = 20+6 = 26`

`text(Due to symmetry,)`

`text(Pr) (-2<Z<1)` `= text(Pr) (8<X<26)`
  `= text(Pr) (14<X<32)`

`=>   B`

Probability, MET1 2006 VCAA 5

Let `X` be a normally distributed random variable with a mean of 72 and a standard deviation of 8. Let `Z` be the standard normal random variable. Use the result that `text(Pr) (Z < 1) = 0.84`, correct to two decimal places, to find

  1. the probability that `X` is greater than `80`  (1 mark)

  2. the probability that  `64 < X < 72`  (1 mark)

  3. the probability that  `X < 64`  given that  `X < 72`  (2 marks)

Show Answers Only

  1. `0.16`
  2. `0.34`
  3. `8/25`

Show Worked Solution

a.   vcaa-2006-meth-5ai

`text(Pr) (X > 80)`

`= text(Pr) (Z > (80 – 72)/8)`

`= text(Pr) (Z > 1)`

`= 0.16`

 

b.   `text(Pr) (64 < X < 72)`

♦ Mean mark 45%.

`= text(Pr) (72 < X < 80)\ \ \ text(due to symmetry)`

`= 0.5 – 0.16`

`= 0.34`

 

♦ Mean mark 40%.
MARKER’S COMMENT: Notation was poor, showing a lack of understanding in this area.

c.   `text(Conditional probability)`

`text(Pr) (X < 64 | X < 72)` `= (text{Pr} (X < 64))/(text{Pr} (X < 72))`
  `= 0.16/0.50`
  `= 8/25 or 0.32`

Probability, MET1 2012 VCAA 8a

The random variable `X` is normally distributed with mean 100 and standard deviation 4.

If  `text(Pr) (X < 106) = q`,  find  `text(Pr) (94 < X < 100)`  in terms of  `q`.   (2 marks)

Show Answers Only

`q – 1/2`

Show Worked Solution

`text(Given)\ \ text(Pr) (X < 106) = q,`

♦ Mean mark 40%.
MARKER’S COMMENT: Those who drew a diagram to visualise the symmetry were the most successful.

`=> text(Pr) (X < 94) = 1-q`

 

`text(By symmetry of normal curve:)`

`:.\ text(Pr) (94 < X < 100)`

`= 1/2 – (1 – q)`

`= q – 1/2`