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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 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 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)`