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Statistics, 2ADV EQ-Bank 34

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 

  1. Find the median test mark.   (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.
  3. Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.   (2 marks)
  4. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.   (1 mark)

Show Answers Only

i.    `6`

ii.   `text(43%)`

iii.  `text(The statement assumes the data is normally distributed which is incorrect.)`

Show Worked Solution
♦ Mean mark (i) 50%.
♦♦Mean mark (ii) 34%.

i.    `text(Median)= text(15th + 16th score)/2= (4 + 8)/2= 6`
 

ii.   `text(Lower limit) = 5.4-4.22 = 1.18`

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`
 

iii.   `text(The statement assumes the data is normally distributed.)`

♦♦♦ Mean mark (iii) 13%.

`text(This is incorrect in this case.)`

Statistics, 2ADV EQ-Bank 31

In a particular country, the birth weight of babies is normally distributed with a mean of 3000 grams. It is known that 95% of these babies have a birth weight between 1600 grams and 4400 grams.

One of these babies has a birth weight of 3497 grams. What is the `z`-score of this baby's birth weight?   (2 marks)

Show Answers Only

`0.71`

Show Worked Solution

`text(95% babies within)\ 1600-4400\ \text{grams:}`

`3000+2\sigma` `=4400`
`2sigma` `= 1400`
`sigma` `= 1400/2=700`

 
`:. ztext(-score)\ (3497)= (x-mu)/\sigma= (3497-3000)/700= 0.71`

Statistics, 2ADV EQ-Bank 29

The results of two tests are normally distributed. The mean and standard deviation for each test are displayed in the table.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \text{Mathematics} & \quad \text{English} \quad \\
\hline
\rule{0pt}{2.5ex} \quad \mu \quad \rule[-1ex]{0pt}{0pt} & 70 & 75 \\
\hline
\rule{0pt}{2.5ex} \quad \sigma \rule[-1ex]{0pt}{0pt} & 6.5 & 8 \\
\hline
\end{array}

Kristoff scored 74 in Mathematics and 80 in English. He claims that he has performed better in English.

Is Kristoff correct? Justify your answer using appropriate calculations.   (2 marks)

Show Answers Only

`text(He is correct.)`

Show Worked Solution

`text(In Maths:)`

♦ Mean mark 44%.

`ztext{-score (74)}= (x-mu)/sigma= (74-70)/6.5= 0.6153…`

`text(In English:)`

`ztext{-score (80)}=(80-75)/8= 0.625`
 

`text(Kristoff’s)\ ztext(-score in English is higher than his)\ z text(-score in Maths.)`

`:.\ text(He is correct. He performed better in English.)`

Statistics, 2ADV EQ-Bank 28

A machine produces nails. When the machine is set correctly, the lengths of the nails are normally distributed with a mean of  6.000 cm  and a standard deviation of  0.040 cm.

To confirm the setting of the machine, three nails are randomly selected. In one sample the lengths are  5.950,  5.983 and  6.140.

The setting of the machine needs to be checked when the lengths of two or more nails in a sample lie more than 1 standard deviation from the mean.

Does the setting on the machine need to be checked? Justify your answer with suitable calculations.   (2 marks)

Show Answers Only

`mu = 6.000,\ \sigma = 0.040`

`text{Limits for}\ ±1 \sigma:`

`6.000 + 0.040 = 6.040\ text{(upper)}`

`6.000-0.040 = 5.960\ text{(lower)}`

`text(Chosen nails:)\ \ 5.950` `=>\ text(outside limits)`
`5.983` `=>\ text(inside)`
`6.140` `=>\ text(outside)`

 

`text{Since 2 nails are outside 1}\ \sigma\ \text{limits}`

`=>\ \ \text{Settings need to be checked.}`

Show Worked Solution

`mu = 6.000,\ \sigma = 0.040`

`text{Limits for}\ ±1 \sigma:`

`6.000 + 0.040 = 6.040\ text{(upper)}`

`6.000-0.040 = 5.960\ text{(lower)}`

`text(Chosen nails:)\ \ 5.950` `=>\ text(outside limits)`
`5.983` `=>\ text(inside)`
`6.140` `=>\ text(outside)`

 

`text{Since 2 nails are outside 1}\ \sigma\ \text{limits}`

`=>\ \ \text{Settings need to be checked.}`

Statistics, 2ADV EQ-Bank 13

The formula to calculate `z`-scores can be rearranged to give

`mu = x-\sigma z`

 

where    `mu` is the mean
  `x` is the score
  `sigma` is the standard deviation
  `z` is the `z`-score
  1. In an examination, Aaron achieved a score of 88, which corresponds to a `z`-score of 2.4.
  2. Substitute these values into the rearranged formula above to form an equation.   (1 mark)
  3. In the same examination, Brock achieved a score of 52, which corresponds to a `z`-score of  –1.2.
  4. Using this information, form another equation and solve it simultaneously with the equation from part (a) to find the values of `mu` and `\sigma`.   (2 marks)
Show Answers Only

a.    `mu = 88-2.4\sigma`

b.    `64`

Show Worked Solution

a.   `mu = 88-2.4\sigma`
 

b.   `mu = 52 + 1.2\sigma\ …\ (1)`

`mu = 88-2.4\sigma \ …\ (2)`
 

`text(Subtract)\ \ (2)-(1):`

`0= 36-3.6\sigma\ \ =>\ \ \sigma= 10`

 
`text(Substitute)\ \ \sigma = 10\ \ text(into)\ (1):`

`mu= 52 + 1.2 xx 10= 64`

Statistics, 2ADV EQ-Bank 20

The weights of boxes of Brekky Bicks are normally distributed. The mean is 754 grams and the standard deviation is 2 grams.

  1. What is the `z`-score of a box of Brekky Bicks with a weight of 754 g?   (1 mark)

  2. What is the weight of a box that has a `z`-score of  –1?   (1 mark)

  3. Brekky Bicks boxes are labelled as having a weight of 750 g. What percentage of boxes will have a weight less than 750 g?   (2 marks)

Show Answers Only

a.    `0\ text{(mean)}`

b.    `752\ text(grams)`

c.    `text(2.5%)`

Show Worked Solution

a.    `text{z-score (754 g) = 0  (754 g is the mean)}`

 

b.    `ztext(-score)` `= (x-mu)/sigma`
`-1` `= (x-54)/2`
 `x-754` `= -2`
`x`  `= 752\ text(grams)`

 
c.    `text{z-score (750)= (750-754)/2= -2`

 

`:.\ text(Graph shows that 2.5% of boxes will weigh less than 750 g.)`

Statistics, 2ADV EQ-Bank 19

Two brands of light bulbs are being compared. For each brand, the life of the light bulbs, in hours, is normally distributed and described in the table below.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \quad \text{Mean} \quad & \text{Standard Deviation} \\
\hline
\rule{0pt}{2.5ex} \text{Brand A} \rule[-1ex]{0pt}{0pt} & 450 & 25 \\
\hline
\rule{0pt}{2.5ex} \text{Brand B} \rule[-1ex]{0pt}{0pt} & 500 & 50 \\
\hline
\end{array}

  1. One of the Brand B light bulbs has a life of 400 hours. 
  2. What is the `z`-score of the life of this light bulb?   (1 mark)

  3. A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:
  4. ‘Brand A light bulbs are more likely to be defective than Brand B light bulbs.’
  5. Is this claim correct? Justify your answer, with reference to `z`-scores or standard deviations or the normal distribution.   (2 marks)
Show Answers Only

a.    `-2`

b.    `text(The claim is incorrect.)`

Show Worked Solution

a.    `z text{-score of Brand B bulb (400 hrs)}`

`= (x-mu)/sigma= (400-500)/50= -2`
 

b.   `z text{-score of Brand A bulb (400 hours)} =(400-450)/25=-2`

`text(S)text(ince the)\ z text(-score for both brands is –2, they are equally likely to be defective.)`

`:.\ text(The claim is incorrect.)`

Statistics, 2ADV EQ-Bank 16

The results of two class tests are normally distributed. The means and standard deviations of the tests are displayed in the table.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \quad \text{Test 1} \quad & \quad \text{Test 2} \quad \\
\hline
\rule{0pt}{2.5ex} \text{Mean} \rule[-1ex]{0pt}{0pt} & 60 & 58 \\
\hline
\rule{0pt}{2.5ex} \text{Standard Deviation} \rule[-1ex]{0pt}{0pt} & 6.2 & 6.0 \\
\hline
\end{array}

  1. Stuart scored 63 in Test 1 and 62 in Test 2. He thinks that he has performed better in Test 1. Do you agree? Justify your answer using appropriate calculations.   (2 marks)

  2. If 150 students sat for Test 2, how many students would you expect to have scored less than 64?   (2 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `126`

Show Worked Solution

a.    `text(In Test 1:)\ \ mu = 60,\ sigma = 6.2`

`z text(-score)\ (63)= (x-mu)/sigma= (63-60)/6.2= 0.483…`

 
`text(In Test 2:)\ \ mu = 58,\ sigma = 6.0`

`z text(-score)\ (62)= (62-58)/6.0=0.666…`

`text(S) text(ince Stuart’s)\ z\ text(-score is higher in Test 2, his performance relative)`

`text(to the class is better despite his mark being slightly lower.)`
 

b.    `text(In Test 2:)`

`z text(-score)\ (64)= (64-58)/6= 1`

`text(84% have)\ z text(-score) < 1`

`:.\ text(# Students expected below 64) = text(84%) xx 150 = 126`

Statistics, 2ADV EQ-Bank 9 MC

The scores on an examination are normally distributed with a mean of 70 and a standard deviation of 6. Michael received a score on the examination between the lower quartile and the upper quartile of the scores.

Which shaded region most accurately represents where Michael's score lies?
 

A.
 		 B.
 
C. D.
Show Answers Only

`A`

Show Worked Solution

`text{68% of marks lie between 64 and 76 (mean ± 1 σ).}`

`text(50% of marks lie between)\ Q_1\ text(and)\ Q_3.`

`=> A`

Statistics, 2ADV EQ-Bank 5 MC

A machine produces cylindrical pipes. The mean of the diameters of the pipes is 8 cm and the standard deviation is 0.04 cm.

Assuming a normal distribution, what percentage of cylindrical pipes produced will have a diameter less than 7.96 cm?

  1. `text(16%)`
  2. `text(32%)`
  3. `text(34%)`
  4. `text(68%)`
Show Answers Only

`A`

Show Worked Solution

`mu = 8\ text(cm)\ \ \ s = 0.04\ text(cm)`

`ztext{-score(7.96)}= (x-mu)/sigma= (7.96-8)/0.04=-1`
 

2UG 2015 20MC Answer

`:.\ text(% of pipes with a diameter less than 7.96 cm.)`

`=\ text(50%)-text(34%)`

`=\ text(16%)`

`=>A`

Statistics, 2ADV EQ-Bank 7 MC

The weights of  10 000 newborn babies in NSW are normally distributed. These weights have a mean of 3.1 kg and a standard deviation of 0.35 kg.

How many of these newborn babies have a weight between 2.75 kg and 4.15 kg?

  1. `4985`
  2. `6570`
  3. `8370`
  4. `8385`
Show Answers Only

`D`

Show Worked Solution

`text(Find)\ z text(-scores of 2.75 and 4.15 kg)`

`z\ (2.75)` `= (x-mu)/5 = (2.75-3.1)/0.35 = -1`
`z\ (4.15)` `= (4.15-3.1)/0.35 = 3`

 
`text(68% between)\ z=–1\ text(and 1)\ => \ text(34% between)\ z=–1\ text(and 0)`

`text(99.7% between)\ z=–3\ text(and 3) \ => \ text(49.85% between)\ z=0\ text(and)\ 3`
 

`%\ text(with)\ z text(-scores between)\ –1\ text(and 3) = 34 + 49.85 =\ text(83.85%)`

`text(Babies between 2.75 kg and 4.15 kg) = text(83.85%) xx 10\ 000 = 8385`

`=>  D`

Statistics, 2ADV EQ-Bank 6 MC

Which of these graphs best represents positively skewed data with the smaller standard deviation?
 

2UG-2006-8abMC

2UG-2006-8cdMC

Show Answers Only

`C`

Show Worked Solution

`text(By elimination:)`

`text(Positive skew when the tail on the right side is longer.)`

`:.\ text(NOT)\ B\ text(or)\ D`

`text(A smaller standard deviation occurs when data is clustered more closely.)`

`:.\ text(NOT)\ A\ text(where data is more widely spread.)`

`=>  C`

Statistics, 2ADV EQ-Bank 4 MC

The pulse rates of a large group of 18-year-old students are approximately normally distributed with a mean of 75 beats/minute and a standard deviation of 11 beats/minute. 

The percentage of 18-year-old students with pulse rates less than 53 beats/minute or greater than 86 beats/minute is closest to

  1. `2.5text(%)` 
  2. `5text(%)` 
  3. `16text(%)` 
  4. `18.5text(%)` 
Show Answers Only

`D`

Show Worked Solution

`mu=75,\ \ \ sigma=11`

`z text{-score (53)}=(x-mu) /sigma=(53-75)/11= -2`

`z text{-score (86)}= (86-75)/11=1`

core 2008 VCAA 6-7

`text{% of students }(z<-2\ ∪ \ z>1) =2.5+16=18.5 text(%)`

 `=>D`

Statistics, 2ADV EQ-Bank 37

The probability density function for the normal distribution with mean \(\mu\) and standard deviation \(\sigma\) is

\(f(x)=\dfrac{1}{\sigma \sqrt{2 \pi}} e^{-\tfrac{(x-\mu)^2} {2 \sigma^2}}\)

The graph of  \(y=e^{-\tfrac{1}{2}(x-1.5)^2}\)  is shown. The point \(M\) is a local maximum.
 

Using a \(z\)-score table of values, calculate the area of the shaded region. Give your answer correct to three decimal places.   (4 marks)

Show Answers Only

\(\text{Comparing the graph to the normal distribution PDF:}\)

\(\mu=1.5, \ \sigma=1\)

\(e^{-\tfrac{1}{2}(x-1.5)^2} = \sqrt{2 \pi} \times f(x)\)

\(\Rightarrow\ \text{Total area under the curve} = \sqrt{2 \pi}\ \text{u}^2\)
 

\(\text{Convert the \(x\)-values to \(z\)-scores:}\)

\(\text{When }\ x=0:\ \ z=\dfrac{0-1.5}{1}=-1.5 \)

\(\text{When }\ x=1.5:\ \ z=\dfrac{1.5-1.5}{1}=0 \)

\(P(Z \leqslant 0)=0.5000\)

\(P(Z \leqslant 1.5)=0.9332\ \ \text{(from table)}\)

\(P(Z \leqslant -1.5)=1-0.9332=0.0668\ \ \text{(by symmetry)}\)

\(P(-1.5 \leqslant Z \leqslant 0)=0.5000-0.0668=0.4332\)
 

\(\text{Area under curve} = 0.4332 \times \sqrt{2\pi} \approx 1.08587\)

\(\text {Shaded area}\) \(=\ \text{Area of rectangle}-\text{Area under curve}\)
  \(=(1.5 \times 1)-1.08587 \ldots\)
  \(=0.414 \ \text{u}^2 \ \text{(3 d.p.)}\)
Show Worked Solution

\(\text{Comparing the graph to the normal distribution PDF:}\)

\(\mu=1.5, \ \sigma=1\)

\(e^{-\tfrac{1}{2}(x-1.5)^2} = \sqrt{2 \pi} \times f(x)\)

\(\Rightarrow\ \text{total area under the curve} = \sqrt{2 \pi}\ \text{u}^2\)
 

\(\text{Convert the \(x\)-values to \(z\)-scores:}\)

\(\text{When }\ x=0:\ \ z=\dfrac{0-1.5}{1}=-1.5 \)

\(\text{When }\ x=1.5:\ \ z=\dfrac{1.5-1.5}{1}=0 \)

\(P(Z \leqslant 0)=0.5000\)

\(P(Z \leqslant 1.5)=0.9332\ \ \text{(from table)}\)

\(P(Z \leqslant -1.5)=1-0.9332=0.0668\ \ \text{(by symmetry)}\)

\(P(-1.5 \leqslant Z \leqslant 0)=0.5000-0.0668=0.4332\)
 

\(\text{Area under curve} = 0.4332 \times \sqrt{2\pi} \approx 1.08587\)

\(\text {Shaded area}\) \(=\ \text{Area of rectangle}-\text{Area under curve}\)
  \(=(1.5 \times 1)-1.08587 \ldots\)
  \(=0.414 \ \text{u}^2 \ \text{(3 d.p.)}\)

Statistics, 2ADV S3 2025 HSC 23

  1. In a flock of 12 600 sheep, the ratio of males to females is \(1:20\).
  2. The weights of the male sheep are normally distributed with a mean of 76.2 kg and a standard deviation of 6.8 kg.
  3. In the flock, 15 of the male sheep each weigh more than \(x\) kg. 
  4. Find the value of \(x\).   (4 marks)

  5. The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the weights of male sheeр.
  6. Explain whether it could be expected that 300 of the females from the flock each weigh more than \(x\) kg, where \(x\) is the value found in part (a).   (1 mark)

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Show Worked Solution

a.    \(12\,600 \ \text{sheep} \ \Rightarrow \ \text{male : female}=1:20\)

\(\text{Number of sheep in “1 part”} = \dfrac{12\,600}{21}=600\)

\(\Rightarrow \ \text{male : female}=600:12\,000\)

\(\text{male sheep} \ \%=\dfrac{15}{600}=0.025\%\)

\(z \text{-score }(0.025 \%)=2\)

\(\text{Using } \ z=\dfrac{x-\bar{x}}{s}, \ \text{find} \ \ x_m:\)

\(2\) \(=\dfrac{x_m-76.2}{6.8}\)  
\(x_m\) \(=76.2+2 \times 6.8=89.8 \ \text{kg}\)  

 
b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that 300 females weigh > 89.8 kg.}\)

♦♦ Mean mark (b) 35%.

Statistics, 2ADV S3 2025 HSC 8 MC

The minimum daily temperature, in degrees, of a town each year follows a normal distribution with its mean equal to its standard deviation. The minimum daily temperature was recorded over one year.

What percentage of the recorded minimum daily temperatures was above zero degrees?

  1. 16%
  2. 50%
  3. 68%
  4. 84%
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider a possible example:}\)

\(\text{Let mean min daily temperature = 8°C}\)

\(\text{Std dev = 8°C}\)

\(z\text{-score (0°C)}\ =-1\)

\(\text{Percentage above 0°C} = 50+34=84\%\)

\(\Rightarrow D\)

♦ Mean mark 41%.

Statistics, 2ADV S3 2024 HSC 23

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)
 

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

Statistics, 2ADV S3 2024 HSC 3 MC

Pia's marks in Year 10 assessments are shown. The scores for each subject were normally distributed.

\begin{array}{|l|c|c|c|}
\hline & \textit {Pia's mark} & \textit {Year 10 mean} & \textit {Year 10 standard} \\
&&&\textit {deviation}\\
\hline \text {English} & 78 & 66 & 6 \\
\hline \text {Mathematics} & 80 & 71 & 10 \\
\hline \text {Science} & 77 & 70 & 15 \\
\hline \text {History} & 85 & 72 & 9 \\
\hline
\end{array}

In which subject did Pia perform best in comparison with the rest of Year 10?

  1. English
  2. Mathematics
  3. Science
  4. History
Show Answers Only

\(A\)

Show Worked Solution

\(\text {Consider the z-score of each option:}\)

\(z \text {-score (English)}=\dfrac{78-66}{6}=2\)

\(z \text {-score (Maths)}=\dfrac{80-71}{10}=0.9\)

\(z \text {-score (Science) }=\dfrac{77-70}{15}=0.46 \ldots\)

\(z \text {-score (History})=\dfrac{85-72}{9}=1.4 \ldots\)

\(\Rightarrow A\)

Statistics, 2ADV S3 2023 HSC 23

A random variable is normally distributed with a mean of 0 and a standard deviation of 1 . The table gives the probability that this random variable lies below `z` for some positive values of `z`.

The probability values given in the table are represented by the shaded area in the following diagram.
 

The weights of adult male koalas form a normal distribution with mean `mu` = 10.40 kg, and standard deviation `sigma` = 1.15 kg.

In a group of 400 adult male koalas, how many would be expected to weigh more than 11.93 kg?  (4 marks)

Show Answers Only

`37\ text{koalas*}`

`text{*36 or 36.72 koalas would also receive full marks}`

Show Worked Solution
`ztext{-score (11.93)}` `=(x-mu)/sigma`  
  `=(11.93-10.4)/1.15`  
  `=1.330`  

 
`Ptext{(Koala weighs > 11.93 kg)}\ = P(z>1.330)`

`text{Using the table:}`

`P(z>1.33)` `=1-0.9082`  
  `=0.0918`  

 

`:.\ text{Expected koalas > 11.93 kg}` `=0.0918 xx 400`  
  `=36.72`  
  `=37\ text{koalas*}`  

 
`text{*36 or 36.72 koalas would also receive full marks}`

Statistics, 2ADV S3 2022 HSC 26

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definintion)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`

Statistics, 2ADV S3 EQ-Bank 2 MC

The heights of females living in a small country town are normally distributed:

    • 16% of the females are more than 160 cm tall.
    • 2.5% of the females are less than 115 cm tall.

The mean and the standard deviation of this female population, in centimetres, are closest to

  1. mean = 135               standard deviation = 15
  2. mean = 135               standard deviation = 25
  3. mean = 145               standard deviation = 15
  4. mean = 145               standard deviation = 20
Show Answers Only

`C`

Show Worked Solution

`160 -> ztext{-score} = 1`

`115 -> ztext{-score} = -2`

`1` `= {160 – mu}/sigma`
`mu + sigma` `= 160\ …\ (1)`
`-2` `= {115-mu}/sigma`
`mu-2 sigma` `= 115\ …\ (2)`

 
`(1)-(2)`

`3sigma` `=45`  
`sigma` `=15`  

 
`text{Substitute}\ \ sigma = 15\ \ text{into (1)}`

`mu = 145` 

`=> C`

Statistics, 2ADV S3 2021 HSC 32

In a particular city, the heights of adult females and the heights of adult males are each normally distributed.

Information relating to two females from that city is given in Table 1.
 

The means and standard deviations of adult females and males, in centimetres, are given in Table 2.
 


 

A selected male is taller than 84% of the population of adult males in this city.

By first labelling the normal distribution curve below with the heights of the two females given in Table 1, calculate the height of the selected male, in centimetres, correct to two decimal places.  (4 marks)

 

Show Answers Only

`178.95 \ text{cm}`

Show Worked Solution

 

`z text{-score (175 cm, female)} = 2`

♦ Mean mark 41%.

`z text{-score (160.6 cm, female)} = -1`
 

`text{Find} \ mu \ text{of female heights:}`

`mu – sigma` `= 160.6`  
`mu + 2sigma` `= 175`  
`3 sigma` `= 175 – 160.6`  
`sigma` `= 14.4/3`  
  `= 4.8 \ text{cm}`  
`:. \ mu` `= 165.4 \ text{cm}`  

 

`text{Selected male’s height has} \ z text{-score} = 1`

`mu text{(male)} = 1.05 times 165.4 = 173.67`

`sigma \ text{(male)} = 1.1 times 4.8 = 5.28`

 

`:. \ text{Actual male height}` `= 173.67 + 5.28`  
  `= 178.95 \ text{cm}`  

Statistics, 2ADV S3 2021 HSC 22

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable lies between 0 and `z` for different values of `z`.
 

   

The probability values given in the table for different values of `z` are represented by the shaded area in the following diagram.
 

  1. Using the table, find the probability that a value from a random variable that is normally distributed with a mean of 0 and standard deviation 1 lies between 0.1 and 0.5.  (1 mark)

  2. Birth weights are normally distributed with a mean of 3300 grams and a standard deviation of 570 grams. By first calculating a `z`-score, find how many babies, out of 1000 born, are expected to have a birth weight greater than 3528 grams.  (3 marks)

Show Answers Only
  1. `0.1517`
  2. `345\ text(babies)`
Show Worked Solution

♦♦ Mean mark part (a) 29%.
COMMENT: Note the Advanced and Std2 questions varied slightly but used the same table and graph.
a.    `P(0.1 < x < 0.5)` `= 0.1915 – 0.0398`
    `= 0.1517`
 

b.   `mu = 330, sigma = 570`

♦ Mean mark part (b) 48%.
`ztext(-score)\ (3528)` `= (x – mu)/sigma`
  `= (3528 – 3300)/570`
  `= 0.4`

 

`P(ztext(-score) > 0.4)`  `= 0.5 – 0.1554`
  `= 0.3446`

 
`:.\ text(Expected babies > 3528 grams)`

`= 1000 xx 0.3446`

`= 344.6`

`~~ 345\ text(babies)`

Statistics, 2ADV S3 2020 HSC 28

In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5.

  1. Two adults who both work are chosen at random.

     

    Find the probability that at least one of them earns between $15 and $30 per hour.  (3 marks)

  2. The number of adults who work is equal to three times the number of adults who do not work.

     

    One adult is chosen at random.

     

    Find the probability that the chosen adult works and earn more than $25 per hour.  (2 marks)

Show Answers Only
  1. `0.965775`
  2. `3/8`
Show Worked Solution

`ztext(-score)\ ($15) = (x – mu)/sigma = (15 – 25)/5 = −2`

♦ Mean mark part (a) 40%.

`ztext(-score)\ ($30) = (30 -25)/5 = 1`
 

`text(Percentage of scores where)\  −2 <= z <= 1`

`= 81.5text(%)`
 

`P(text(at least one earns between $15 – $30))`

`= 1 – P(text(neither))`

`= 1 – (1-0.815)^2`

`=1-0.185^2`

`= 0.965775`


b.   `P(text(works)) = 3/4, \ P(text(earns) > $25) = 1/2`

`:. P(text(works and earns) > $25)`

`= 3/4 xx 1/2`

`= 3/8`

Statistics, 2ADV S3 2020 HSC 9 MC

Suppose the weight of melons is normally distributed with a mean of `mu` and a standard deviation of `sigma`.

A melon has a weight below the lower quartile of the distribution but NOT in the bottom 10% of the distribution.

Which of the following most accurately represents the region in which the weight of this melon lies?
 

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`text(Distributions using)\ mu and sigma:`
 

 
`text(Adjusting the above to identify the interval  10% < weight < 25%)`

  
`=>C`

Statistics, 2ADV S3 2020 HSC 3 MC

John recently did a class test in each of three subjects. The class scores on each test were normally distributed.

The table shows the subjects and John's scores as well as the mean and standard deviation of the class scores on each test.
 

 
Relative to the rest of class, which row of the table below shows John's strongest subject and his weakest subject?
 

Show Answers Only

`A`

Show Worked Solution

`text(Calculate the)\ ztext(-score of each subject:)`

`ztext{-score (French)} = frac(82 – 70)(8) = 1.5`

`ztext{-score (Commerce)} = frac(80 – 65)(5) = 3.0`

`ztext{-score (Music)} = frac(74 – 50)(12) = 2.0`

 
`therefore \ text{Commerce is strongest, French is weakest}`

`=> \ A`