Normal distribution - z scores.
Test Yourself 1.
| On this page, questions address: |
| 1. Calculating each of the four important numbers. |
| 2. Interpreting z scores. |
| 3. Using z scores to combine your marks. |
| Calculating z scores. | 1. Ish scores 62 in a test where the mean is 48 and the standard deviation is 6.
What is her z score?
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2. Jean-Pierre scores 48 in a test where the mean is 52 and the standard deviation is 8.
What is his z score? Answer.z = -0.5 |
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| 3. Jelena scores 51 in a test where the mean is 50 and the standard deviation is 4.
What is her z score? Answer.z = 0.25 |
4. Brett scores 42 in a test where the mean is 55 and the standard deviation is 5.
What is his z score? Answer.z = -2.6 |
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| Calculating the mean or the SD. | 5. Lindsay calculates her z-score to be 1.85 after receiving her assessment task result of 68%. The
results on the task were normally distributed.
If the standard deviation on the task was 8.5, what was the mean? Answer.Mean = 52.28 |
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| 6. The length of ribbons made each day is normally distributed with a mean length of 50 m. One roll of ribbon is measured and its length was 49.8 which gave a z score of -0.75. What is the standard deviation (i.e. the variability) of the ribbons being manufactured? Answer.SD = 0.27 m or 27 cm. |
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| 7. A batch of Covid-19 vaccine is made to precise levels but even so there are slight variations. The tolerance in the variability requires a standard deviation of 0.135 ml across a batch of 1,000 vials.
One vial is tested and it has 4.85 ml which gives that vial a z-score of -1.112. What is the mean amount of vaccine in one vial |
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| 8. When my Charles was born, he was 14 days beyond the mean gestation time for a pregnancy of 280 days (and he has been late ever since). Later that day (for his birth day), I calculated his z score as being 2.8.
What is the standard deviation for a normal gestation period |
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| 9. sdz=0 | ||||||||||||||||||||||||||||||||||||||||||||
| Calculating a value. | 10. The class mean for a maths assessment was 55 and the standard deviation was 10.
What mark did Seb receive if we know his z score was -1.5.
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11. An assessment task had a mean of 62.5% and a standard deviation of 9%. |
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| 12. A set of assay results to determine whether a mine should be opened showed that the distribution of gold across 36 samples had a mean of 2.76 grams/ton and the standard deviation across all samples was 0.85.
One sample had a z score of 2.64. What was the number of grams/ton for that sample (answer to 2 decimal places)? Answer.Assay was 5.00 grams/ton. |
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| 13. The length of each of 20 beetles collected on a field trip was measured. The mode length was 2.5 cm, the median length was 2.35 cm and the mean length was 2.26 cm. The standard deviation of the lengths of the 20 beetles was 0.4 cm.
The researcher determined that the z score of one beetle was 1.85. How was the actual length of that beetle? Answer.Actual length was 3 cm. |
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14. Jörg was given his score in a Trial exam as -0.7 (to 1 decimal place). The Trial had a mean score of 63 and a standard deviation of 5.6. What was Jörg's actual score? Answer.Actual score was 59. |
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| Interpreting z scores. | 15. | |||||||||||||||||||||||||||||||||||||||||||
17.
Answer.(i) 1 ≤ z ≤ 2. (ii) % = 47.5% - 34% = 13.5%. |
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18. Answer.(i) A z score of -1 means that the length of the component is 10 - 0.02 = 9.98 cm. (ii) A component in section A is between 10 and 10.02 cm say 10.01 cm. (iii) Number = 500 × (47.5%-34%) = 500×13.5% i.e. 68 components approx. |
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| Using z scores to calculate total marks. | 20. The table below contains marks obtained by Damien and Alyssa in their English and Mathematics assessments.
Also included are the means and standard deviations for both subjects (as calculated using Excel).
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