Exercises  The Normal Curve


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1 Exercises  The Normal Curve 1. Find e following proportions under e Normal curve: a) P(z>2.05) b) P(z>2.5) c) P(1.25<z<0.54) d) P(z>3.6) 2. Suppose e life span of e common fruit fly is normally distributed wi average 18.3 days and standard deviation 6.3 days. a) How likely is it at a randomly selected fruit fly will live 21 days or longer? b) If a fruit fly outlived 99% of all oer flies, how long did it live? c) How many flies live between 7 and 14 days (inclusive)? 3. Suppose a class of 150 students wrote an exam. The exam grades were normally distributed wi a mean grade of 67% and a standard deviation of 8.5%. a) You attained a grade equivalent to a zscore of z=0.7. What grade did you receive? b) Compute e minimum grade required to be in e class 84 percentile. 4. The number of hours at fulltime employees work per week at a large Nor American corporation has a mean of 52.1 hours per week and a standard deviation of 3.1 hours per week. Assume e data is normally distributed. a) What fraction of employees work between 45.9 hours and 58.3 hours? b) What percentage of employees work longer an 60 hours? c) Due to drastic budget cuts, e entire first quartile of is data set (i.e. e bottom 25% of employees in terms of working hours) is to be eliminated. What is e minimal number of hours tyou need to be working each week in order to not lose your job? 5. The average price of a new PC is $949. If computer prices are normally distributed wi standard deviation $100, what proportion of computers cost over $1200? What is e interquartile range of is data set? 6. Exam scores are normally distributed. If a student wi a 75% is in e 86 percentile, and a student wi a 55% is in e 14 percentile, find e mean and standard deviation of exam scores. 7. Exam score are normally distributed wi mean 72% and standard deviation 12%. Sort e following grades from highest to lowest: i) An 80% ii) The 80 percentile iii) A zscore of z=0.8 iv) The mode 8. A researcher takes weight samples of a mice population, and finds at e mice weights are normally distributed wi average 145 grams and standard deviation 21 grams. However, e researcher en realizes at his/her scale was not adjusted correctly, and added 12 grams to every object weighed (e.g. an object weighing 100 grams would display as 112 grams). Describe e real weight distribution of e mice? Is it still normal? What is its mean and standard deviation now?
2 9. A website owner finds at 72% of visitors use Internet Explorer. What is e probability at less an 550 out of his last 800 visitors used Internet Explorer. 10. Roughly 1 in 12 males have some form of colour blindness. a) What is e probability at among a group of one hundred males between 10 and 20 have a form of colour blindness? b) What is e probability at among a group of one ousand males more an 100 have some form of colour blindness? Exercises  The Sampling Distribution 1. Assume at e height of Canadian women aged is normally distributed wi =158 cm kg and = 13 cm. a) How likely is it at a randomly selected woman will be at most 150 cm tall? b) If a sample of 50 women is selected, find e probability at e sample s mean height is between 155 cm and 160 cm. 2. Grades in a very large class of Stat 100 students have average 72% and standard deviation 13%. If 40 students were selected at random, what is e probability at eir average grade exceeds 70%? 3. A lightbulb is advertised to last on average 3000 hours. Suppose at, in fact, lightbulb lifetimes are normally distributed wi an average of 2800 hours and standard deviation 190 hours. a) What is e probability at a randomly selected lightbulb will burn 3000 hours? b) What is e probability at ten bulbs, burning consecutively, will last 30,000 hours? c) If you purchase a pack of six lightbulbs, what is e probability at at least two of em will burn for 3000 hours?
3 NORMAL CURVE  ANSWERS: 1 a) P(z>2.05) = =.0202 b) P(z>2.5) = = c) P(1.25<z<0.54) = =.189 d) Note: is is no longer on e table. Any zscore value beyond 3.5 is automatically assigned a highest proportion of Similarly, any zscore below 3.5 is given proportion (To get exact answers, you would need eier a different table or use computer software). Hence, for our class, e answer is P(z>3.6)= = a) x=21 days converts to z=0.43. Look up A= The probablity at a fruit fly lives 21 days or longer is.3336 (or 33.36%) b) Look up A=.99 to find a zscore of z=2.33 (would also accept halfway between 2.32 and 2.33). Convert z=2.33 to xscore: x= (6.3)(2.33) = This fly lived about 33 days. c) Convert x=7 to z=1.79, look up A= Convert x=14 to z=.68, look up A= About =.2116 (or 21.16%) of flies live between 7 and 14 days. 3. a) Convert z back to x=x +(z)(s) = 67 + (0.7)(8.5) = Your score is 73%. b) To be in e 84 percentile, we need e proportion of data to our left to be.84. From Table A, at corresponds to a zscore between 0.99 and 1.00, say z= Hence you must score at least 67+(0.995)(9.5) = 75.46, or at least a 76% (round up!). 4. a) Convert x=45.9 to z=2 (look up in table, corresponds to proportion of ) and x=58.3 to z=+2 (look up in table, proportion.9772). Hence e proportion of employees working between 45.9 and 58.3 hours is =.9544 b) Convert x=60 to z=2.55. Look up a proportion of Hence e proportion who work more an 60 hours is =.0054 (just over ½ percent). c) Look up a proportion of.25 in table, get zscore of Convert z=.675 to x=(52.1)(.675)(3.1)= hours. You need to be working at least 50 hours a week. 5. Convert x=1200 to z=2.51, look up proportion.994 in table. Hence e proportion of computers costing over $1200 is =.006 To find e interquartile range, look up proportions 0.25 (z=.675) and.75 (z=+.675), and convert to prices x=$ and x=$ The interquartile range is $ $ = $135.
4 6. Note at 86 and 14 percentile are mirror images of each oer. Look up zscores for A=.86 to find z=1.08, look up A=.14 to find z= The middle (z=0, i.e. e mean) is exactly halfway in between, hence mean = (75%+55%)/2 = 65% Since a 75% (10% above e mean of 65%) is z=1.08 standard deviations above e mean, e standard deviation must be 10%/1.08 = 9.3% 7. Convert all to equivalent measures, e.g. e actual grade x. i) This is already x=80% ii) Look up A=.8 to find z=0.84, convert to x=72+(12)(0.84)=82.08 iii) Convert z=.8 to x=72 + (12)(.8)=81.6 iv) The mean is equal to e mean (and median) in a normal distribution, so x=72. In order (from highest to lowest): (ii), (iii), (i), (iv) 8. Since each xvalue needs to be decreased by 12 grams, e entire density curve will shift to e left on e xaxis. The shape of e curve remains e same (i.e. e distribution is still normal), e new mean will be 133 grams, e standard deviation remains e same at 21 grams. 9. We have repeated experiments (wi n=800), success (p=.72) or failure (q=.28) and independent trials, so is is binomial. However, calculating P(X<550) is cumbersome. Check if e normal approximation can be used: np = 576 >10, nq = 224 >10. YES! Convert to a normal curve wi mean =np = 576 and = npq = and continuous correction factor X<549.5 Hence P(X<550) = P( Z < ( ) / 12.70) = P (Z < 2.09) =.0183 There is just a 1.83% chance at fewer an 550 visitors used Internet Explorer. 10. This is binomial, wi p=1/12 and q=11/12. Again, we will use e normal approximation. a) Use =np = 8.33 and = npq = 2.76 Find P(10.5<X<19.5) = P(0.79<Z<4.05) = =.2147 (note: z=4.05 is off e chart, hence use A=.9999) b) Use =np = and = npq = 8.74 P(X>100.5) = P ( Z > 1.96) = =.025
5 SAMPLING DISTRIBUTION  ANSWERS 1. a) We want P(X<150). Since X (e height) is normally distributed, we can use e normal table. Convert X=150 to z=( )/13 = Look up an area of A= Hence e probability of a woman being at most 150 cm tall is b) This is now a sampling distribution. Since e sample size is larger an 30 we can assume a normal distribution for wi =158 and =13/ 50=1.84. (Note: since e original distribution is normal, we actually don t need at large a sample). We now want P(155< 160). Convert to zscores. X=155 becomes z=1.63 and X=160 becomes z=1.09. Look up bo values: z=1.63 yields A=.0516 and z=1.09 yields A= Hence e probability of selecting a sample of 50 women wi average height between 155 cm and 160 cm is = Note: since e sample size is large enough, we can assume at e sampling distribution is normal (even ough e original distribution is likely not) wi =72 and =13/ 40=2.06. Convert X=70 to z=.97. Look up z=.97 to find A= Hence e probability of e sample average exceeding 70% is = a) Find P(X>3000) ( at least is implied). P(X>3000)=P(z>1.05) = b) Note: saying at ten lightbulbs will burn consecutively for 30,000 hours is e same as saying at a sample of ten lightbulbs will burn a mean of = 3,000 hours each. Hence is is a sampling distribution of size n=10. Since e original distribution is normal, e sampling distribution for even such a small sample is normal, wi =2800 and =190/ 10=60 hours. Find P( >3000)=P(z>3.33)= = The probability is c) Careful! This is a binomial distribution (repeated trials wi n=6, success/failure wi success being burn for 3000 hours wi p=.1469 from part (a), and independent trials). Hence we want e binomial probability, 6 5 P(X 2) = 1  P(X=0)  P(X=1) = 1  (.8531)  6 (.1469)(.8531) = The probability at at least two of six lightbulbs will last 3000 hours is.2163.
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