Normal distrib

The questions on this page are grouped in categories:

1. Areas under the normal curve - basic percentages.

2. Areas under the normal curve - using tables.

3. Determining expected numbers of cases and percentages.

4. Using only 1 tail.

5. Using two tails.

6. mixed questions.

Basic area under the normal curve.	1. In a normal distribution, which three measures are all equal? (a) the mean, median and skew. (b) the mode, standard deviation and the skew? (c) the mean, median and the mode? (d) the standard deviation and the range. Answer.(c) .
	2. What percentage of the total area under the normal curve falls: (i) between z = 0 and z = 2? (ii) between z = -1 and z = + 2? (iii) below z = -2? (iv) above z = +1?
Area under the normal curve using tables.	6. Using tables of the normal distriution, determine the percentage of the total area which falls: (i) below z = -1.5. (ii) below z = -2.5. (iii) above z = -1.7. (iv) above z = + 2.1.
	7. Using tables of the normal distriution, determine the percentage of the total area which falls between: (i) z = -3 and z = -2? (ii) z = +1 and z = +2? (iii) z = - 2.4 and z = +1.6? (iv) z = -0.5 and + 0.5?
	8. An outlier is commonly defined as being a score: below Q1 - 1.5× interquartile range; or above Q3 + 1.5× interquartile range. What percentage of the normal distribution lies between these limits?
	9. What z score cuts off: (i) the top 10% of the total area under a normal distribution? (ii) the lowest 20% of the total area under a normal distribution?
	10.
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) Summer covers December to February. How many summer days in Cummins Range would you expect the temperature 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.
	15. The weight of tennis balls is normally distributed with a mean of 57.5 grams and a standard deviation of 0.5 grams. (i) What is the range in weights allowed for one of the four Grand Slam events if the ball is to have a 99% chance of being in that range? (ii) What percentage of tennis balls weigh between 57.5 grams and 58 grams?
	16.
1 tail >	17. 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?
	18. 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. The heights of 500 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.420 students.
	20. In a normal distribution, 400 people have a z score less than -2. How many people in this population? Answer.N = 16,000.
2 tail even	21. 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.
	22. In a normal distribution, 850 people have a z score between -1 and +1. How many people in this population? Answer.N = 1,250.
2 tail uneven	23. Light bulbs have a mean operational life of 6,000 hours. The standard deviation of their operational life is 135 hours. Their operational life is normally distributed. From one batch, 6,194 light bulbs had an operational life of between 5,865 hours and 6,270 hours. How many light bulbs were in this batch? Answer.N = 7,600 light bulbs.
	24. . 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.
	25. 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?
	The minimum amount of tread that car tyres can legally have is 1.6 mm. A new tyre will have about 8 mm or more of tread. A survey conducted by a leading automobile association recently measured tyre tread depth on 1200 cars across several localities. The survey showed that the mean depth was 5 mm and the standard deviation was 1.6 mm. (i) What percentage of tyres are expeced to have tread greater than 8.0 mm? (ii) What percentage of tyres will be approaching a tyre replacement because the depth is between 1.6 mm and 3.2 mm? (iii)What percentage of tyres on these cars were illegal?

	(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?
	The average length of an adult Grey Nurse shark is 3.2 m while the standard deviation of the adult population is 0.4 m. (i) What percentage of the Grey Nurse adult population is between 2.8 m and 3.6 m in length? (ii) What percentage of Grey Sharks is longer than a tall swimmer who is 2 m in height? (iii) A swimmer had a close encounter with a Grey Nurse while surf-boarding. He reported that the shark was at least 4½ m long. How would you react to this news?