Acme Confectionery has launched a new bar called Chocolate Crunch. The weights of these new bars are normally distributed with a mean of 4-64 g and a standard deviation of 0-12 g. A sample of 10 bars is selected at random and the mean weight of the sample is found. Find the probability that the mean weight of the sample is between 4-6 g and 4-7 g.

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Acme Confectionery has launched a new bar called Chocolate Crunch - Leaving Cert Mathematics - Question 8 (i) - 2018

Question 8 (i)

Acme Confectionery has launched a new bar called Chocolate Crunch. The weights of these new bars are normally distributed with a mean of 4-64 g and a standard deviat... show full transcript

Worked Solution & Example Answer:Acme Confectionery has launched a new bar called Chocolate Crunch - Leaving Cert Mathematics - Question 8 (i) - 2018

Step 1

Calculate $z_1$ for 4-6 g

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Answer

To find $z_1$ , we use the formula:

$z_1 = \frac{X - \mu}{\sigma / \sqrt{n}}$

Where:

$X = 4.6$ g,
$\mu = 4.64$ g,
$\sigma = 0.12$ g,
$n = 10$ .

Substituting, we get:

$z_1 = \frac{4.6 - 4.64}{0.12 / \sqrt{10}} = \frac{-0.04}{0.0379} \approx -1.05409$ .

Step 2

Calculate $z_2$ for 4-7 g

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Answer

For $z_2$ , we apply the same formula:

$z_2 = \frac{X - \mu}{\sigma / \sqrt{n}}$

Where:

$X = 4.7$ g.

Substituting, we get:

$z_2 = \frac{4.7 - 4.64}{0.12 / \sqrt{10}} = \frac{0.06}{0.0379} \approx 1.581138$ .

Step 3

Find the probability $p(-1.05 < z < 1.58)$

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Answer

We can find the probability by using the standard normal distribution table:

From the Z-table, $P(Z < 1.58) \approx 0.9429$ and $P(Z < -1.05) \approx 0.1487$ .

Thus, the required probability is:

$p(-1.05 < z < 1.58) = P(Z < 1.58) - P(Z < -1.05) = 0.9429 - 0.1487 = 0.7942$ .

This approximates to 0.796 or 79.6%.

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