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Solutions to problem 4.3, 4.9, 4.19, 4.24, 4.19, and 4.29 from stats 13.1 homework 3. It includes the setup, formulas to use, and calculations to determine probabilities of certain events based on normal distributions.
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Setup: ฮผ = 1400 ฯ = 100 โ Y โผ N (1400, 100) Formulas to use: z = yโฯ ฮผ P (Y 6 y) = P (z 6 yโฯฮผ )
P (Y 6 1500) = P (z 6 1500100 โ^1400 ) = P (z 6 1 .00) =. 8413 From Table 3 in the text or SOCR the area is .8413 or 84.13% This means that 84.13% of the data (Swedish male brains) is below 1500 grams.
= P (z 6 1500100 โ^1400 ) โ P (z 6 1325100 โ^1400 ) = P (z 6 1 .00) โ P (z 6 โ.75) =. 8413 โ .2266 =. 6147 or 61 .47% 61.47% of the area is between 1325g and 1500g.
We want to know P (Y > 1325) Since we already know P (Y 6 1325) = P (z 6 โ.75) = .2266 we can simply P (Y > 1325) = 1 โ P (Y 6 1325) = 1 โ .2266 = .7734 77.34% of the data is above 1325g.
P (Y > 1475) โ z = 1475100 โ^1400 =. 75 Note that this is the mirror image (75g over rather than 75g under the mean) of a previous question where we learned that 22.66% of the data is below 1325g. Also note that the z-score is .75 rather than -.75. Therefore our hypothesis is that 22.66% of the data is above 1475g because the normal distribution is symmetrical. If using Table 3 we find that the area below 1475 is .7734. Thus, the percentage greater than 1475 is 1-.7734 = .2266.
2 Problem 4.
Setup: ฮผ = 176mg/dLi ฯ = 30mg/dLi โ Y โผ N (176, 30) Formulas to use: z = yโฯ ฮผ P (Y 6 y) = P (z 6 yโฯฮผ )
P (Y > 186) = 1 โ P (Y 6 186) = 1 โ P (z 6 18630 โ^176 ) = 1 โ P (z 6 .33) = 1 โ. 6293
P (Y 6 156) = P (z 6 15630 โ^176 ) = P (z 6 โ.67) =. 2514
P (Y 6 216) = P (z 6 21630 โ^176 ) = P (z 6 1 .33) =. 9082
P (Y > 121) = 1 โ P (Y 6 121) = 1 โ P (z 6 12130 โ^176 ) = 1 โ P (z 6 โ 1 .83) = 1 โ .0336 =. 9664
Since we already know the z-scores for both 186 and 216 and the associated percentages of data below both: P (z 6 1 .33) โ P (z 6 .33) =. 9082 โ .6293 =. 2789
Since we already know the z-scores for both 121 and 156 and the associated percentages of data below both:
. 2514 โ .0336 =. 2178
Since we already know the z-scores for both 156 and 186 and the associated percentages of data below both:
. 6293 โ .2514 =. 3779
3 Problem 4.
To determine if a data set is normally distributed a Q-Q plot is used. This is created by plotting the Rainfall data score on the y-axis and the z-score (also called the Normal Score) on the x-axis. If the data is normally distributed the data will be aligned on the y = x line. The Q-Q plot below is slightly curved, indicating the distribution of the sample is skewed to the right. It does not appear that the precipitation distribution is normal.
4 Problem 4.
Setup: ฮผ = 7. 8 ฯ = 2. 3 โ Y โผ N (7. 8 , 2 .3) As the data is discrete (whole numbers) rather than continuous and the Normal distribution
P (Y 6 120) = P (z 6 12022 โ^145 ) = P (z 6 โ 1 .14) =. 1271 12.71% of corn plants are shorter than 120 cm.
We already know the z-score and corresponding area for 120 cm. z = 15022 โ^145 =. 23 From Table 3 the area below .23 is .5910. .5910-.1271 =. 46.39% of corn plant are between 120 cm and 150 cm in height.
We already know the z-score and corresponding area for 100 cm and 120 cm.
. 1271 โ .0202 =. 1069
z = 18022 โ^145 = 1. 59 From Table 3, the area to the left of 1.59 is .9441, but we are interested in the area to the right (greater than). So we subtract from 1: 1-.9441 =.0559โ 5.59% of corn plants are taller than 180 cm.
We already know the z-score and corresponding area for 150 cm, .23 and .5910 respectively. Therefore, 59.10% of corn plants are less than 150 cm.
6 Problem 4.
Setup: ฮผ = 7. 3 ฯ = 11. 1 โ N (7. 3 , 11 .1)
P (Y > 10) = 1 โ P (Y < 10) = 1 โ P (z < 1011 โ.^71. 3 ) = 1 โ P (z < .24) = 1 โ .9207 =. 40.52% of the change in heart rates is above 10 beats per minute.
P (Y > 20) = 1 โ P (Y < 20) = 1 โ P (z < 2011 โ.^71. 3 ) = 1 โ P (z < 1 .14) = 1 โ .8729 =. 1271 12.71% of the change in heart rates is above 20 beats per minute.
P (5 6 Y 6 15) = P (Y 6 15) โ P (Y 6 5) = P (z 6 1511 โ.^71. 3 ) โ P (z 6 511 โ^7. 1.^3 ) = P (z 6 .69) โ P (z 6 โ.21) =. 7549 โ .4168 = .3381 33.81% of the change in heart rates is between 5 and 15 beats per minutes.
