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Statistics, STD2 S5 2024 HSC 5 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, STD2 S5 2021 HSC 41

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 16%.

`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, STD2 S5 2020 HSC 35

The intelligence Quotient (IQ) scores for adults in City A are normally distributed with a mean of 108 and a standard deviation of 10.

The IQ score for adults in City B are normally distributed with a mean of 112 and a standard deviation of 16.

  1. Yin is an adult who lives in City A and has an IQ score of 128.
    What percentage of the adults in this city have an IQ score higher than Yin's?   (2 marks) 

  2. There are 1 000 000 adults living in City B.
    Calculate the number of adults in City B that would be expected to have an IQ score lower than Yin's.  (2 marks)

  3. Simon, an adult who lives in City A, moves to City B. The  `z` -score corresponding to his IQ score in City A is the same as the `z`-score corresponding to his IQ score in City B.
    By first forming an equation, calculate Simon's IQ score. Give your answer correct to one decimal place.   (3 marks)

Show Answers Only
  1. `text(2.5%)`
  2. `840 \ 000`
  3. `101.3`
Show Worked Solution

a.    `text{In City A:}`

♦ Mean mark 50%.

`z text{-score}\ (128) = frac {x -mu}{sigma} = frac{128 – 108}{10} = 2`
 


 

`therefore\ text{2.5% have a higher IQ in City} \ A.`
 

b.    `text{In City B:}`

♦ Mean mark 44%.

`z text{-score}\ (128) = frac{128 – 112}{16} = 1`
 


 

`therefore \ text{Adults in City} \ B \ text{with an IQ}\ <  128`

`= 84text(%) xx 1 \ 000 \ 000`

`= 840 \ 000`
 

c.    `z text{-score in City A}\ = z text{-score in City B}`

♦♦♦ Mean mark 17%.
`frac{x – 108}{10} ` `= frac{x -112}{16} \ \ text{(multiply b.s.} \ xx 160 text{)}`
`16 (x – 108)` `= 10 (x – 112)`
`16 x – 1728` `= 10 x – 1120`
`6x` `= 608`
`x` `= 101.3`

  
`therefore \ text{Simon’s IQ} = 101.3 \ text{(to 1 d.p.)}`

Statistics, STD2 S5 2020 HSC 8 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`

Statistics, STD2 S5 2018 HSC 27e

Joanna sits a Physics test and a Biology test.

  1. Joanna’s mark in the Physics test is 70. The mean mark for this test is 58 and the standard deviation is 8.

     

    Calculate the `z`-score for Joanna’s mark in this test.  (1 mark)

  2. In the Biology test, the mean mark is 64 and the standard deviation is 10.

     

    Joanna’s `z`-score is the same in both the Physics test and the Biology test.

     

    What is her mark in the Biology test?  (2 marks)

Show Answers Only
  1. `1.5`
  2. `79`
Show Worked Solution

i.   `x = 70, \ mu = 58, \ sigma = 8`

`:. ztext(-score)` `= (x – mu)/sigma`
  `= (70 – 58)/8`
  `= 1.5`

 

ii.    `1.5` `= (x – 64)/10`
  `x – 64` `= 15`
  `:. x` `= 79`

Statistics, STD2 S5 2017 HSC 29d

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.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. 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
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

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

♦♦ Mean mark 34%.

`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)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

Statistics, STD2 S5 2015 HSC 28b

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

2015 28b

 
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)`

`text(his)\ z text(-score in Maths.)`

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

Statistics, STD2 S5 2007 HSC 25d

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

 

  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
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `126`
Show Worked Solution

i.  `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,)`

`text(his performance relative to the class is better)`

`text(despite his mark being slightly lower.)`

 

ii.  `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, STD2 S5 2011 HSC 27c

Two brands of light bulbs are being compared. For each brand, the life of the light bulbs is normally distributed.

2011 27c

  1. One of the Brand B light bulbs has a life of 400 hours. 

     

    What is the  `z`-score of the life of this light bulb?  (1 mark)

  2. A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:

     

    ‘Brand A light bulbs are more likely to be defective than Brand B light bulbs.’

     

    Is this claim correct? Justify your answer, with reference to  `z`-scores or standard deviations or the normal distribution.   (2 marks)

Show Answers Only
  1. `-2`
  2. `text(The claim is incorrect.)`
Show Worked Solution

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

`= (x-mu)/sigma`
`= (400\-500)/50`
`= –2`

 

ii.   `z text{-score of Brand A bulb (400 hours)}`

♦ Mean mark 42%
MARKER’S COMMENT: A number of students found drawing normal curves in their solution advantageous.
`=(400-450)/25`
`=–2`

 
`text(S)text(ince the)\ z text(-score for both brands is –2,)`

`text(they are equally likely to be defective.)`

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

Statistics, STD2 S5 2013 HSC 29b

Ali’s class sits two Geography tests. The results of her class on the first Geography test are shown.

`58,\ \ 74,\ \ 65,\ \ 66,\ \ 73,\ \ 71,\ \ 72,\ \ 74,\ \ 62,\ \ 70`

The mean was 68.5 for the first test. 

  1. Calculate the standard deviation for the first test. Give your answer correct to one decimal place.    (1 mark)

  2. On the second Geography test, the mean for the class was 74.4 and the standard deviation was 12.4.

     

    Ali scored 62 on the first test. Calculate the mark that she needed to obtain in the second test to ensure that her performance relative to the class was maintained.   (3 marks)

Show Answers Only
  1. `5.2\ \ \ text{(to 1 d.p.)}`
  2. `text(Ali needs to score 58.9)`
Show Worked Solution
♦ Mean mark 39%
COMMENT: Make sure you are confident with this function on your calculator!
i. `sigma` `=5.2201…`
    `=5.2`  `text{(to 1 d.p.)}`

 

ii. `z =(x-mu)/sigma`
`z text{-score (1st test)}` `= (62-68.5)/5.2`
  `=-1.25`

 
`text(2nd test has)\ z text(-score of)\-1.25 :`

♦♦ Mean mark 21%.
MARKER’S COMMENT: When “performance relative to the class is maintained”, `z text(-scores)` are the same in each test.
`-1.25` `= (x-74.4)/12.4`
`x-74.4` `=-15.5`
`x` `=58.9`

 

`:.\ text(Ali needs to score 58.9)`