Aussie Maths & Science Teachers: Save your time with SmarterEd
The lifetime of a certain brand of batteries has a mean lifetime of 20 hours and a standard deviation of 2 hours. A random sample of 40 batteries is selected.
The probability that the mean lifetime of this sample of 40 batteries exceeds 19.5 hours is closest to
\(D\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)
\(\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 20, \dfrac{4}{40}\right)\)
\(\sigma_{\bar{X}} = \dfrac{1}{\sqrt{10}} \)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\overline{X}-20}{\frac{1}{\sqrt{10}}}\sim N(0,1)\)
\(Z=\dfrac{19.5-20}{\frac{1}{\sqrt{10}}}=-1.58 \ \text{(2 d.p.)}\)
\(\text{Using Normal Distribution Table of Values:}\)
\(\Pr(\overline{X}>19.5)=\Pr(Z>-1.58)\)
\(\phantom{\Pr(\overline{X}>19.5)}=0.9429\)
\(\Rightarrow D\)
The waiting time, \(T\) hours, to see a particular doctor at a clinic has a mean of 0.5 hours and a standard deviation of 0.3 hours.
A sample of 35 waiting times is chosen at random.
Use the standard normal distribution table (provided) to find the probability that the average waiting time of the sample of 35 patients is between 0.43 hours and 0.50 hours. Give your answer as a percentage correct to 1 decimal place. (3 marks)
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\(\Pr(0.43<\overline{T}<0.50) \approx 41.6\%\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)
\(\text{The sample mean,}\ \overline{T}, \text{for random samples of size 35 is}\)
\(\text{approximately normally distributed, where:}\)
\(\mu=0.5\ \ \text{and}\ \ \sigma=\dfrac{0.3}{\sqrt{n}}=\dfrac{0.3}{\sqrt{35}} \approx 0.0507\)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\overline{T}-0.5}{\frac{0.3}{\sqrt{35}}} \sim N(0,1)\)
| \(\Pr(0.43<\overline{T}<0.50)\) | \(=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)\) |
| \(=\Pr(-1.38<Z<0)\) | |
| \(=\Pr(0<Z<1.38)\) | |
| \(=0.9162-0.5000\) | |
| \(=0.4162 = 41.6\%\ \text{(1 d.p.)}\) |
A company accountant has previously found that the mean amount owed on any individual unpaid invoice is \(\$800\) with a standard deviation of \(\$200\).
Using the normal distribution table (included), determine the probability that, in a random sample of 60 unpaid invoices, the total amount owed is more than \(\$47\,500\). Give your answer as a percentage correct to one decimal place. (3 marks)
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\(62.6\%\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)
\(\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n} \right) \sim N\left(800, \dfrac{200^2}{60} \right)\)
\(P(\Sigma X) > 47\,500 = P\left(\bar{X}>\dfrac{47\,500}{60}\right)=P(\bar{X})>791.67 \)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\bar{X}-800}{\frac{200}{\sqrt{60}}}\sim N(0,1)\)
\(Z=\dfrac{791.67-800}{\frac{200}{\sqrt{60}}}=-0.32 \ \text{(2 d.p.)}\)
\(\text{Using Normal Distribution Table of Values:}\)
| \(P(\Sigma X > 47\,500)\) | \(=P(Z>-0.32)\) |
| \(=1-P(Z\leq-0.32)\) | |
| \(=1-0.3745\) | |
| \(=62.6\%\) |
The gestation period of cats has a mean of 66 days and a variance of 9 days\(^2\).
A sample of 32 cats is chosen at random.
Using the Normal Distribution Table of Values, determine the probability that the sample has an average gestation period greater than 65 days. (3 marks)
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\(0.9706\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)
\(\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 66, \dfrac{9}{32}\right)\)
\(\sigma_{\bar{X}} = \dfrac{3}{\sqrt{32}} \)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\overline{X}-66}{\frac{3}{\sqrt{32}}}\sim N(0,1)\)
\(Z=\dfrac{65-66}{\frac{3}{\sqrt{32}}}=-1.89 \ \text{(2 d.p.)}\)
\(\text{Using Normal Distribution Table of Values:}\)
| \(P(\overline{X}>65)\) | \(=P(Z>-1.89)\) |
| \(=1-P(Z\leq -1.89)\) | |
| \(=1-0.0294\) | |
| \(=0.9706\) |
The weights of pumpkins on a particular farm are normally distributed with a mean of 13.5 kg and a standard deviation of 3 kg.
A random sample of 15 pumpkins is selected.
Let \(X_1, X_2, X_3, \ldots, X_{15}\) denote the weights of the pumpkins in the sample, and let
\(\overline{X}=\dfrac{X_1+X_2+X_3+\cdots+X_{15}}{15}\).
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a. \(13.5\)
b. \(0.775\)
a. \(E(\overline{X})=\mu=13.5\)
b. \(\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{3}{\sqrt{15}} \approx 0.775\)
A research team is investigating the amount of sleep obtained by Year 12 students. Previous studies indicate that the sleep time of Year 12 students has a population mean of 7.4 hours and a population standard deviation of 2.42 hours.
A random sample of 100 Year 12 students is selected.
Using the normal distribution table (included), determine the probability that the mean sleep time of the sample is less than 7 hours. Give your answer correct to four decimal places. (3 marks)
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\(4.94 \%\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)
\(\text{The sample mean,}\ \overline{X}, \text{for random samples of size 100 is}\)
\(\text{approximately normally distributed, where:}\)
\(\mu=7.4\ \ \text{and}\ \ \sigma=\dfrac{2.42}{\sqrt{n}}=\dfrac{2.42}{\sqrt{100}}\)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\overline{X}-7.4}{\frac{2.42}{\sqrt{100}}} \sim N(0,1)\)
\(Z=\dfrac{7-7.4}{\frac{2.42}{\sqrt{100}}}=-1.65 \ \text{(2 d.p.)}\)
\(\text{Using Normal Distribution Table of Values:}\)
\(P(\overline{X} < 7)=P(Z < -1.65)=0.0494=4.94 \%\)
In a large school, the average amount of money spent per student per day at the canteen is $8 with a standard deviation of 6.5 .
At the end of each day, 50 randomly chosen students are asked how much they spent at the canteen on that day.
Use the standard normal distribution table (included) to find the probability that the sample mean on a particular day is greater than $10. (3 marks)
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\(1.46 \%\)
\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}\)
\(\text{The sample mean,}\ \overline{X}, \text{for random samples of size 50 is}\)
\(\text{approximately normally distributed, where:}\)
\(\mu=8\ \ \text{and}\ \ \sigma=\dfrac{6.5}{\sqrt{n}}=\dfrac{6.5}{\sqrt{50}}\)
\(\text{By the central limit theorem:}\)
\(Z=\dfrac{\overline{X}-8}{\frac{6.5}{\sqrt{50}}} \sim N(0,1)\)
\(Z=\dfrac{10-8}{\frac{6.5}{\sqrt{50}}}=2.18 \ \text{(2 d.p.)}\)
\(\text{Using Normal Distribution Table of Values:}\)
| \(P(\overline{X} \geq 10)\) | \(=1-P(Z \leq 2.18)\) |
| \(=1-0.9854\) | |
| \(=1.46 \%\) |