Normal Distribution Activity (solutions)

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Practice Demo: Suppose that 33 percet of wome believe i the existece of alies. If 100 wome are selected at radom, what is the probability that more tha 45 percet of them will say that they believe i alies? Role #1: 100 wome selected 45 percet of them Role #2: µ ( p ˆ ) p SD( ) p(1 p) 0.33(1 0.33) SD ˆp 0.04702 100 Role #3: µ ( ) 0.33 ( ) Role #2:.189.236.283.33.377.424.471 Proportio of 100 wome that believe i alies.189.236.283.33.377.424.471 Proportio of 100 wome that believe i alies ormalcdf(.45, 1E99, 0.33, 0.04702 ) 0.00535

1. Suppose family icomes i a tow are ormally distributed with a mea of $1,200 ad a stadard deviatio of $600 per moth. What is the probability that a family has a icome betwee $1,400 ad $2,250? Role #1: X No sample of size greater tha oe was take. Family icomes are the populatio. Role #2: µ σ Role #3: µ 1200 σ 600-600 0 600 1200 1800 2400 3000 Family icome i $ X -600 0 600 1200 1800 2400 3000 Family icome i $ X ormalcdf( 1400, 2250, 1200, 600 ) 0.3294

2. A opiio poll asks, Are you afraid to go outside at ight withi a mile of your home because of crime? Suppose that the proportio of all adults who would say Yes to this questio is 0.4. Assume that the poll obtaied 20 aswers radomly. What percet of such polls with 20 resposes have 10 or more say Yes. Role #1: 20 resposes 10 or more say Yes proportio of all adults Role #2: µ ( p ˆ ) p SD( ) p(1 p) 0.4(1 0.4) SD ˆp 0.10954 20 Role #3: µ ( ) 0.4 ( ).071.181.290.4.510.619.729 Proportio of 20 adults that are afraid to go outside at ight Havig 10 or more out of a sample of 20 say Yes is equivalet to havig 10 p ˆ. I other words, p ˆ 0. 5. 20.071.181.290.4.510.619.729 Proportio of 20 adults that are afraid to go outside at ight ormalcdf( 0.5, 1E99, 0.4, 0.10954 ) 0.180655 or about 18.1% of such polls

3. Fid the area uder the curve betwee the z-scores of -2 ad 1. Role #1: Z No sample of size greater tha oe was take. z-scores Role #2: µ 0 σ 1 Role #3: µ 0 σ 1-3 -2-1 0 1 2 3 Z-scores Z -3-2 -1 0 1 2 3 Z-scores Z ormalcdf( -2, 1, 0, 1 ) 0.81859

4. Adult ose legth is ormally distributed with mea 45mm ad stadard deviatio 6mm. Fid the probability that the sample mea ose legth is betwee 44mm ad 46mm for radom samples of 36 adults. Role #1: x samples of 36 adults sample mea ose legth Role #2: µ ( x ) µ SD( x) σ 6 Role #3: µ ( x) 45 SD( x) 1 36 42 43 44 45 46 47 48 Mea ose legth (mm) of 36 adults x 42 43 44 45 46 47 48 Mea ose legth (mm) of 36 adults x By the Empirical Rule we see that the aswer is about 68% because 44mm ad 46mm is exactly oe stadard deviatio each way o the sample mea ose legth distributio ( x distributio). More precisely we have ormalcdf( 44, 46, 45, 1 ) 0.68269 or 68.269%

5. The weight of a particular brad of cookies has a ormal distributio with a mea weight of 32 ouces ad a stadard deviatio of 0.3 ouces. Whe we look at the mea weight of 20 packages, 68% of them will be betwee what two values? Role #1: x mea weight of 20 packages σ Role #2: µ ( x ) µ SD( x) 0. 3 Role #3: µ ( x) 32 SD( x) 0. 06708 20 31.80 31.87 31.93 32 32.07 32.13 32.20 Mea weight (ouces) of 20 packages of cookies x x 31.80 31.87 31.93 32 32.07 32.13 32.20 Mea weight (ouces) of 20 packages of cookies Similar to the previous problem where we used the Empirical Rule, we see that the aswer is: Whe we look at the mea weight of 20 packages ( x values) about 68% of them will be betwee 31.93 ouces ad 32.07 ouces.

6. A restaurateur aticipates servig about 180 people o a Friday eveig, ad believes that about 20% of the patros will order the chef s steak special. How may of those meals should he pla o servig i order to be pretty sure of havig eough steaks o had to meet customer demad? Justify your aswer, icludig a explaatio of what pretty sure meas to you. Role #1: servig about 180 people 20% of the patros Role #2: µ ( p ˆ ) p SD( ) p(1 p) 0.2(1 0.2) SD ˆp 0. 02981 180 Role #3: µ ( ) 0.2 ( ).111.140.170.2.230.260.289 Proportio of 180 people that order the chef s steak special o a Friday eveig Here the populatio is all patros that eat at that particular restaurat o Friday ights. The 180 people o this Friday eveig is a sample (although ot SRS!). The proportio of those 180 people orderig the chef s steak special is the sample proportio or value. What could this value be? Accordig to the Empirical Rule, we kow that about 99.7% of all values occur betwee 0.111 ad 0.289. It is highly ulikely that is greater tha 0.289 sice this happes oly about 0.15% of the time ( 0.3% 0.15% ). Therefore, we would expect that the 2 proportio of the 180 patros that order the chef s steak would be o more tha 0.289. Sice 29% of 180 people is 52.2 people, we coclude that the restaurateur should pla o servig 53 of those meals. That way the restaurateur ca be pretty sure that orders of chef s steak o Friday eveigs ca be filled (about 99.7% of Friday eveigs).

7. I this example we will be iterested i the heights of orther Europea males. We take such a perso ad reduce him to a sigle umber via the usual operatios for measurig someoe's height. The we model the height of orther Europea males as a ormal populatio with mu 150 cm ad sigma 30 cm. If we sample oe orther Europea male, what's the probability that his height will fall outside of 140 ad 170? I other words, what are the chaces that he'll be either below 140, or he'll be above 170 i height? That's what we mea by the word "outside." Role #1: X sample oe orther Europea Norther Europea males are the populatio. Role #2: µ σ Role #3: µ 150 σ 30 60 90 120 150 180 210 240 Height (cm) of a orther Europea male X 60 90 120 150 180 210 240 Height (cm) of a orther Europea male ormalcdf( -1E99, 140, 150, 30 ) 0.36944 ormalcdf( 170, 1E99, 150, 30 ) 0.25249 The probability that his height will fall outside 140 cm ad 170 cm is 0.36944 + 0.25249 0.62193 or about 62.2% of the time. X

8. Fid the proportio of observatios from the Stadard Normal Distributio which are below 2.45. Role #1: Z No sample of size greater tha oe was take. Stadard Normal Distributio Role #1: µ 0 σ 1 Role #2: µ 0 σ 1 Role #3: -3-2 -1 0 1 2 3 Z-scores Z -3-2 -1 0 1 2 3 Z-scores Z The word proportio is this exercise ca be misleadig sice it is referrig to the percetage (i decimal form) of z-scores less tha 2.45. Here the proportio value is equivalet to the shaded area of the curve. ormalcdf( -1E99, 2.45, 0, 1 ) 0.99286