The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram. It can be seen that the distribution is positively skewed. The mean is €22.05. The median is €17.82. The standard deviation is €10.64. The lower quartile is €12.80. The upper quartile is €26.05. (i) If six employees are selected at random from this population, what is the probability that exactly four of them had hourly earnings of more than €12.80? In a computer simulation, random samples of size 200 are repeatedly selected from this population and the mean of each sample is recorded. A thousand such sample means are recorded. (ii) Describe the expected distribution of these sample means. Your description should refer to the shape of the distribution and to its mean and standard deviation. (iii) How many of the sample means would you expect to be greater than €23?

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The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram - Leaving Cert Mathematics - Question b - 21

Question b

The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram. It can be seen that the distribution is positively skewe... show full transcript

Worked Solution & Example Answer:The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram - Leaving Cert Mathematics - Question b - 21

Step 1

If six employees are selected at random from this population, what is the probability that exactly four of them had hourly earnings of more than €12.80?

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Answer

Since €12.80 is the lower quartile, the probability of any one randomly selected person having hourly earnings of more than €12.80 is 0.75. Using the binomial probability formula, we can express the probability of getting exactly 4 successes (earning more than €12.80) in 6 trials:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

n = 6 (trials)
k = 4 (successes)
p = 0.75 (probability of success)

Thus, $P(4 ext{ successes}) = \binom{6}{4} (0.75)^4 (0.25)^2$ Calculating this:

$\binom{6}{4} = 15$
Then, $= 15 \cdot (0.75)^4 \cdot (0.25)^2$ $= 15 \cdot 0.3164 \cdot 0.0625$ $= 0.296625$

So, the probability is approximately 0.2966.

Step 2

Describe the expected distribution of these sample means. Your description should refer to the shape of the distribution and to its mean and standard deviation.

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Answer

According to the Central Limit Theorem, the distribution of sample means will be approximately normally distributed when the sample size is sufficiently large, regardless of the population's distribution shape. In this case, since we are sampling 200 employees, the distribution of the sample means will approach a normal distribution. The mean of these sample means will be equal to the population mean, which is €22.05.

The standard deviation of the sample means (also known as the standard error) can be calculated using: $\text{Standard Error} = \frac{\sigma}{\sqrt{n}}$ Where:

$\sigma = 10.64$ (population standard deviation)
$n = 200$ (sample size)

Calculating that gives: $\text{Standard Error} = \frac{10.64}{\sqrt{200}} \approx 0.752$

Therefore, the expected distribution of the sample means will be normally distributed with a mean of €22.05 and a standard deviation of approximately €0.752.

Step 3

How many of the sample means would you expect to be greater than €23?

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Answer

Let X be the normally distributed random variable with mean €22.05 and standard deviation approximately €0.752. To find how many of the sample means would be greater than €23, we first find:

$P(X > 23) = 1 - P(X \leq 23)$

We convert €23 to a standard normal variable Z: $Z = \frac{X - \mu}{\sigma} = \frac{23 - 22.05}{0.752} \approx 1.26$

Using Z-tables or probability calculators, we find: $P(Z \leq 1.26) \approx 0.8762$ Thus, $P(X > 23) = 1 - 0.8762 = 0.1238$

Finally, for a thousand samples, the expected number of sample means greater than €23 would be: $1000 \cdot 0.1238 \approx 124$

Therefore, we would expect approximately 124 of the sample means to be greater than €23.

The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram - Leaving Cert Mathematics - Question b - 21

Worked Solution & Example Answer:The distribution of the hourly earnings of all employees in Ireland in October 2009 is shown in the diagram - Leaving Cert Mathematics - Question b - 21

If six employees are selected at random from this population, what is the probability that exactly four of them had hourly earnings of more than €12.80?

Describe the expected distribution of these sample means. Your description should refer to the shape of the distribution and to its mean and standard deviation.

How many of the sample means would you expect to be greater than €23?

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