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In 2015, in a particular country, the weights of 15 year olds were normally distributed with a mean of 63.5 kg and a standard deviation of 10 kg - Leaving Cert Mathematics - Question 8 - 2017

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In 2015, in a particular country, the weights of 15 year olds were normally distributed with a mean of 63.5 kg and a standard deviation of 10 kg. (i) In 2015, Maris... show full transcript

Worked Solution & Example Answer:In 2015, in a particular country, the weights of 15 year olds were normally distributed with a mean of 63.5 kg and a standard deviation of 10 kg - Leaving Cert Mathematics - Question 8 - 2017

Step 1

(i) Find the percentage of 15 year olds in that country who weighed more than Mariska.

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Answer

To find the percentage of 15 year olds who weighed more than Mariska (50 kg), we first need to calculate the z-score:

z=Xμσ=5063.510=1.35z = \frac{X - \mu}{\sigma} = \frac{50 - 63.5}{10} = -1.35

Next, we look up the z-score of -1.35 in the standard normal distribution table, which gives us:

P(Z<1.35)=0.0915P(Z < -1.35) = 0.0915

Therefore, the percentage of 15 year olds who weigh more than Mariska is:

P(Z>1.35)=1P(Z<1.35)=10.0915=0.9085P(Z > -1.35) = 1 - P(Z < -1.35) = 1 - 0.0915 = 0.9085

Hence, approximately 90.85% of 15 year olds weigh more than Mariska.

Step 2

(ii) Find Kamal’s weight.

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Answer

Since 1.5% of 15 year olds are heavier than Kamal, this means that:

P(Z>z)=0.015P(Z > z) = 0.015

Thus, we can find:

P(Z<z)=10.015=0.985P(Z < z) = 1 - 0.015 = 0.985

Looking up in the standard normal distribution table, we find:

z=2.17z = 2.17

Next, we can calculate Kamal's weight using the z-score formula:

z=Xμσz = \frac{X - \mu}{\sigma}

Rearranging gives:

X=zσ+μX = z \cdot \sigma + \mu

Substituting the values:

X=2.1710+63.5=85.2 kgX = 2.17 \cdot 10 + 63.5 = 85.2 \text{ kg}

Thus, Kamal’s weight is approximately 85.2 kg.

Step 3

(iii) State the null hypothesis and your alternative hypothesis.

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Answer

Null Hypothesis (H₀)

The mean weight of 15 year olds has not changed: H0:μ=63.5H_0 : \mu = 63.5

Alternative Hypothesis (H₁)

The mean weight of 15 year olds has changed: H1:μ63.5H_1 : \mu \neq 63.5

To test the hypothesis, we calculate the z-score using the sample statistics:

  • Sample mean (xˉ\bar{x}) = 62 kg
  • Sample size (n) = 150
  • Standard deviation (s) = 10 kg

Using the z-score formula: z=xˉμs/n=6263.510/150=1.8371z = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{62 - 63.5}{10 / \sqrt{150}} = -1.8371

Conclusion

At the 5% significance level, we compare the calculated z-score with critical z-values (-1.96 and 1.96). Since -1.8371 is within -1.96 and 1.96, we fail to reject the null hypothesis. Therefore, there is insufficient evidence to suggest that the mean weight has changed from 2015 to 2016.

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