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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 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 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 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 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 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)
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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 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 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`