Normal distrib - TYS 1 Solns

The questions on this page are grouped in categories:

1. Areas under the normal curve.

3. Determining expected numbers of cases and percentages.

4. Using only 1 tail.

5. Using two tails.

1. Area under the normal curve.	In a normal distribution, which three measures are all equal? (a) the mean, median and skew - incorrect because a mixture of characteristics. (b) the mode, standard deviation and the skew - incorrect because a mixture of characteristics. (c) the mean, median and the mode - correct. All are measures of where the center of the distribution is located and all are the same as the normal curve is symmetrical. (d) the standard deviation and the range - incorrect as both are both measures of spread.

2.




Determining expected numbers and percentages.	12. A set of exam results from 25,000 students is assumed to be normally distributed. How many students from this group are expected to score: (i) up to one standard deviation above the mean? (ii) 2 standard deviations or more below the mean? (iii) between 1 standard deviation below the mean and up to 2 standard deviations above the mean?
	13. Luke completes a statistical analysis on the weights (in kilograms) of each member of his gym. He finds that the data are normally distributed, with a mean of 55.0 and a standard deviation of 6.5. What percentage of members are expected to weigh between 48.5 kg and 68.0 kg? Answer.81.5% of members (34% + 47.5%).
	The maximum daily temperatures during the summer months in Cummins Range, W. A. (where Rare Earth minerals are mined) are recorded. These maximum temperatures are normally distributed with a mean of 21.4°C and a standard deviation of 7.6 °C. (i) What temperature has a z-score of - 1? (ii) What percentage of summer days in Cummins Range would you expect to be between 13.8°C and 36.6°C?
	14. The following graph indicates z-scores of ‘height-for-age’ for girls aged 5–19 years. In each age, height is normally distributed. (i) What is the z-score for a six year old girl of height 120 cm? (ii) Charlotte is 7.5 years old. How tall is Charlotte if just 2½% of girls of the same age are taller than her? (iii) Charlotte remains at her height due to an illness for three years - until she is 10.5 years old. What percentage of girls of the same age will then be taller than Charlotte? (iii)What is the average height of an 16 year old girl? Answer.(i) z = 1. (ii) Height =135 cm. (iii) 84% will be older (z = -1). (iv) An average (z = 0) 16 year old girl is about 162 cm.



1 tail >	The heights of a group of friends are normally distributed with a mean of 160 cm and a standard deviation of 15 cm. What percentage of the group is more than 190 cm tall?
	15. A normal distribution has a mean of 170 and a standard deviation of 10. (i) Jess has a raw score in the shaded region. What could her z-score be? (ii) What percentage of the data lies in the shaded region? Answer.(i)Any value such that 1 ≤ z ≤ 2. (ii) % = 47.5% - 34% = 13.5%.
	19. A height of 169 cm is at a z score of (169 - 172)÷3 = -1. 34% of students are between z = -1 and z = 0 and a further 50% have a z score > 0. Hence a total of 84% The expected number of students with a height in excess of 169 cm is 84% × 500 = 420.
	20.
2 tail even	Heights of surf lifesavers are measured across a large number of clubs. The average height is 178 cm and the standard deviation was 7 cm. The percentage of lifesavers between 164 cm and 192 cm was determined to be 88%. What conclusion can be made about the distribution of the heights of lifesavers? Answer.The heights are not normally distributed because we would expect 95% to be in the range + 2 to -2 standard deviations.

2 tail uneven
	16. The lengths of 500 widgets manufactured at a particular factory are normally distributed with a mean of 10 cm and a standard deviation of 0.02 cm. The diagram below describes the distribution of the lengths of the components about the mean with the z-scores indicated. (i) A component is selected at random and it is measured to have a z score of -1. What is the implication of this measurement? (ii) What is an example of the length of a component which would belong to area A shown in pink? (iii) Approximately how many components will have lengths in the blue Section B? 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.
	17. The heights of 400 students are normally distributed. The mean height is 172 cm and the standard deviation is 3 cm. Find the expected number of students with a height in excess of 169 cm. Answer.>169 = 34% + 50%. &there4:84%×400 = 336 students.


1 tail <	A major seafood chain in NSW and Queensland analyses the hours worked by its staff across all locations. The analysis shows that the average hours worked per week across all permanent and casual staff was 20 hours with a standard deviation of 4 hours. Based on this information and assuming hours worked is normal is normally distributed, what percentage of staff worked less than 12 hours per week?


	(i) 1 ≤ z ≤ 2. (ii) % = 47.5% - 34% = 13.5%.
Mixed questions	The normal distribution below represents the mass of 400 students. All measurements are in kilograms. (i) What is the standard deviation of the distribution? (ii) How many students have a mass in the region marked with an A? (iii) What is the weight of a student with a z-score of - 2 ? (iv) Determine the probability that a student selected at random will have a mass less than 105 kg?