Solved Examples: Normal Distribution

Treat the dataset as a sample
But you may choose to treat it as a population if you wish
(a.)
Empirical Rule: 68 - 95 - 99.7% Rule

$ \bar{x} = 175\;cm \\[3ex] s = 7\;cm \\[3ex] 1s = 1(7) = 7\;cm \\[3ex] 2s = 2(7) = 14\;cm \\[3ex] 3s = 3(7) = 21\; cm \\[3ex] \underline{1\;standard\;\;deviation\;\;below\;\;the\;\;mean} \\[3ex] \bar{x} - 1s \\[3ex] 175 - 7 \\[3ex] 168\;cm \\[3ex] \underline{1\;standard\;\;deviation\;\;above\;\;the\;\;mean} \\[3ex] \bar{x} + 1s \\[3ex] 175 + 7 \\[3ex] 182\;cm \\[3ex] $ This impies that 68% of the basketball team players have heights between 168 cm and 182 cm

$ \underline{2\;standard\;\;deviations\;\;below\;\;the\;\;mean} \\[3ex] \bar{x} - 2s \\[3ex] 175 - 14 \\[3ex] 161\;cm \\[3ex] \underline{2\;standard\;\;deviations\;\;above\;\;the\;\;mean} \\[3ex] \bar{x} + 2s \\[3ex] 175 + 14 \\[3ex] 189\;cm \\[3ex] $ This impies that 95% of the basketball team players have heights between 161 cm and 189 cm

$ \underline{3\;standard\;\;deviations\;\;below\;\;the\;\;mean} \\[3ex] \bar{x} - 3s \\[3ex] 175 - 21 \\[3ex] 154\;cm \\[3ex] \underline{3\;standard\;\;deviations\;\;above\;\;the\;\;mean} \\[3ex] \bar{x} + 3s \\[3ex] 175 + 21 \\[3ex] 196\;cm \\[3ex] $ This impies that 99.7% of the basketball team players have heights between 154 cm and 196 cm

(b.) (i)


$ Center = \bar{x} = 175 \\[3ex] Between\;\;168\;\;and\;\;182 = 68\% \\[3ex] symmetrical \implies \\[3ex] Between\;\;168\;\;and\;\;175 = \dfrac{68}{2} = 34\% \\[5ex] Similarly:\;\;between\;\;175\;\;and\;\;182 = 34\% \\[3ex] Between\;\;161\;\;and\;\;189 = 95\% \\[3ex] symmetrical \implies \\[3ex] Between\;\;161\;\;and\;\;175 = \dfrac{95}{2} = 47.5\% \\[5ex] \implies \\[3ex] Between\;\;161\;\;and\;\;168 = 47.5 - 34 = 13.5\% \\[3ex] Similarly:\;\;between\;\;175\;\;and\;\;189 = 47.5\% \\[3ex] Similarly:\;\;between\;\;182\;\;and\;\;189 = 13.5\% \\[3ex] symmetrical \implies \\[3ex] Between\;\;154\;\;and\;\;196 = \dfrac{99.7}{2} = 49.85\% \\[5ex] \implies \\[3ex] Between\;\;154\;\;and\;\;161 = 49.85 - (34 + 13.5) = 49.85 - 47.5 = 2.35\% \\[3ex] Similarly:\;\;between\;\;175\;\;and\;\;196 = 49.85\% \\[3ex] Similarly:\;\;between\;\;189\;\;and\;\;196 = 2.35\% \\[3ex] Total\;\;area = 100\% \\[3ex] symmetrical \implies \\[3ex] \le 175 = \dfrac{100}{2} = 50\% \\[5ex] \ge 175 = \dfrac{100}{2} = 50\% \\[5ex] \implies \\[3ex] \lt 154 = 50 - (2.35 + 13.5 + 34) = 50 - 49.85 = 0.15\% \\[3ex] Similarly:\;\; \gt 196 = 0.15\% \\[3ex] \lt 161 = 2.35 + 0.15 = 2.5\% \\[3ex] Similarly:\;\; \gt 189 = 2.5\% \\[3ex] \lt 168 = 13.5 + 2.35 + 0.15 = 16\% \\[3ex] Similarly:\;\; \gt 182 = 16\% \\[3ex] Between\;\;161\;\;and\;\;182 = 13.5 + 34 + 34 = 81.5\% \\[3ex] $

Empirical Rule: 68 - 95 - 99.7% Rule
Part of the Empirical Rule states that 99.7% of the adult male tibia bones have lengths within (below and above) 3 standard deviations from the mean

$ \mu = 36.0\; cm \\[3ex] \sigma = 2.8\;cm \\[3ex] 3\sigma = 3(2.8) = 8.4\;cm \\[3ex] \underline{3\;standard\;\;deviations\;\;below\;\;the\;\;mean} \\[3ex] \mu - 3\sigma \\[3ex] 36 - 8.4 \\[3ex] 27.6 \\[3ex] a = 27.6\;cm \\[3ex] \underline{3\;standard\;\;deviations\;\;above\;\;the\;\;mean} \\[3ex] \mu + 3\sigma \\[3ex] 36 + 8.4 \\[3ex] 44.4 \\[3ex] b = 44.4\;cm \\[3ex] $ About 99.7% of adult male tibia bones have lengths between 27.6 cm and 44.4 cm

Empirical Rule: 68 - 95 - 99.7% Rule
Part of the Empirical Rule states that 68% of the adult male tibia bones have lengths within (below and above) 1 standard deviations from the mean

$ \mu = 70 \\[3ex] \sigma = 6 \\[3ex] 1s = 1(6) = 6 \\[3ex] \underline{1\;standard\;\;deviation\;\;below\;\;the\;\;mean} \\[3ex] \mu - 1\sigma \\[3ex] 70 - 6 \\[3ex] 64 \\[3ex] \underline{1\;standard\;\;deviation\;\;above\;\;the\;\;mean} \\[3ex] \mu + 1\sigma \\[3ex] 70 + 6 \\[3ex] 76 \\[3ex] $ This impies that 68% of the examination scores are between 64 and 76
That means that: 34% of the scores are between 64 and 70
Similarly, 34% of the scores are between 70 and 76
But Option B is not the correct answer

One of the properties of the normal curve is that: the mean, median, and mode of a normal curve is the same.
This implies that the median is also 70
median = 70 = 50% of the scores
lower quartile = 25% of the scores = 25% below the median
25% is less than 34%
This implies that 25% below the median is less than 64
upper quartile = 75% of the scores = 25% above the median
Similarly, 25% is less than 34%
This implies that 25% above the median is less than 76
So, we are looking for the curve whose shaded area is less than 64 (from the center to the left) and is also less than 76 (from the center to the right)
This is because 25% is less than 34% (from the center to the left) and 25% is less than 34% (from the center to the right)
This implies that the correct answer is Option A

The question wants us to evaluate Alice's statement using the z-score formula

$ (a.) \\[3ex] \underline{Adult\;\;Female} \\[3ex] \mu = 33.8\;cm \\[3ex] \sigma = 2.2\;cm \\[3ex] x = 34.5 \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{34.5 - 33.8}{2.2} \\[5ex] z = \dfrac{0.7}{2.2} \\[5ex] z = 0.3181818182 \approx 0.32 \\[5ex] \underline{Adult\;\;Male} \\[3ex] \mu = 36\;cm \\[3ex] \sigma = 2.8\;cm \\[3ex] x = 34.5 \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{34.5 - 36}{2.8} \\[5ex] z = -\dfrac{1.5}{2.8} \\[5ex] z = -0.5357142857 \approx -0.54 \\[3ex] $ (b.)
A data value is usual if $-2.00 \lt z-score \lt 2.00$
A data value is unusual if $z-score \lt -2.00$ or if $z-score \gt 2.00$
Adult Female: Because $-2 \lt 0.32 \lt 2$; the data value is usual
Adult Male: Because $-2 \lt -0.54 \lt 2$; the data value is also usual
Both data values are usual.
Hence, there is insufficient evidence to verify Alice' statement.

However,
Because of the positive z-score for the adult female (a positive z-score indicates that the data value is above the mean), the bone may likely be from an adult female than from an adult male.

Let us do the same thing we did for adult male in Question (2.), for adult female.
Empirical Rule: 68 - 95 - 99.7% Rule
Part of the Empirical Rule states that 99.7% of the adult male tibia bones have lengths within (below and above) 3 standard deviations from the mean

$ \mu = 33.8\; cm \\[3ex] \sigma = 2.2\;cm \\[3ex] 3\sigma = 3(2.2) = 6.6\;cm \\[3ex] \underline{3\;standard\;\;deviations\;\;below\;\;the\;\;mean} \\[3ex] \mu - 3\sigma \\[3ex] 33.8 - 6.6 \\[3ex] = 27.2\;cm \\[3ex] \underline{3\;standard\;\;deviations\;\;above\;\;the\;\;mean} \\[3ex] \mu + 3\sigma \\[3ex] 33.8 + 6.6 \\[3ex] = 40.4\;cm \\[3ex] $ About 99.7% of adult female tibia bones have lengths between 27.2 cm and 40.4 cm The option that matches this result is Option (A.)

Empirical Rule: 68 - 95 - 99.7% Rule
Part of the Empirical Rule states that 95% of the data points are within (below and above) 2 standard deviations from the mean.

Based on the Empirical Rule and the Empirical Rule Table above:

$ \mu = 267\;\;days \\[3ex] \sigma = 10\;\;days \\[3ex] \underline{Within\;\;1\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - \sigma = 267 - 10 = 257 \\[3ex] \mu + \sigma = 267 + 10 = 277 \\[3ex] Between\;\;257\;\;and\;\;277 = 68\% \\[3ex] \underline{Within\;\;2\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - 2\sigma = 267 - 2(10) = 267 - 20 = 247 \\[3ex] \mu + 2\sigma = 267 + 2(10) = 267 + 20 = 287 \\[3ex] Between\;\;247\;\;and\;\;287 = 95\% \\[3ex] \underline{Within\;\;3\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - 3\sigma = 267 - 3(10) = 267 - 30 = 237 \\[3ex] \mu + 3\sigma = 267 + 3(10) = 267 + 30 = 297 \\[3ex] Between\;\;237\;\;and\;\;297 = 99.7\% \\[3ex] $ (a.) Mean = 267 days
The percentage of pregnancies last more than 267 days is about 50%

(b.) Between 267 and 277 is 1 standard deviation above the mean
The percentage of pregnancies that last between 267 and 277 days is 34%

(c.) Less than 237 days is less than 3 standard deviations below the mean
The percentage of pregnancies that last less than 237 days is 0.15%

(d.) Between 247 and 287 is 2 standard deviations above the mean
The percentage of pregnancies that last between 247 and 287 days is 95%

(e.) Longer than 287 days is more than 2 standard deviations above the mean
The percentage of pregnancies that last longer than 287 days is 2.35% + 0.15% = 2.5%

(f.) Longer than 297 days is more than 3 standard deviations above the mean
The percentage of pregnancies that last longer than 287 days is 0.15%

$ \mu = 500 \\[3ex] \sigma = 100 \\[3ex] \underline{Within\;\;1\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - \sigma = 500 - 100 = 400 \\[3ex] \mu + \sigma = 500 + 100 = 600 \\[3ex] Between\;\;400\;\;and\;\;600 = 68\% \\[3ex] \underline{Within\;\;2\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - 2\sigma = 500 - 2(100) = 500 - 200 = 300 \\[3ex] \mu + 2\sigma = 500 + 2(100) = 500 + 200 = 700 \\[3ex] Between\;\;300\;\;and\;\;700 = 95\% \\[3ex] \underline{Within\;\;3\;\sigma\;\;from\;\;the\;\;\mu} \\[3ex] \mu - 3\sigma = 500 - 3(100) = 500 - 300 = 200 \\[3ex] \mu + 3\sigma = 500 + 3(100) = 500 + 300 = 800 \\[3ex] Between\;\;200\;\;and\;\;800 = 99.7\% \\[3ex] $ (a.) The test scores that correspond with the provided z-scores is: Option B.



(b.) Scores more than 500 is every score above the mean
The percentage of students that earn quantitative test scores more than 500 is 50%

(c.) Scores between 400 and 600 are scores within 1 standard deviation from the mean
The percentage of students that earn quantitative test scores between 400 and 600 is 68%

(d.) Scores more than 800 are scores more than 3 standard deviations above the mean
The percentage of students that earn quantitative test scores more than 800 is 0.15%

(e.) Scores less than 200 are scores less than 3 standard deviations below the mean
The percentage of students that earn quantitative test scores less than 200 is 0.15%

(f.) Scores between 300 and 700 are scores within 2 standard deviations from the mean
The percentage of students that earn quantitative test scores between 300 and 700 is 95%

(g.) Scores between 700 and 800 are scores in-between 2 standard deviations above the mean and 3 standard deviations above the mean
The percentage of students that earn quantitative test scores between 700 and 800 is 2.35%

(a.) The graph that shows the probability that a z-score is 1.54 or less is option D.


(b.) 1.54 = 1.5 (vertical); 0.04 (horizontal)

First Approach: Left-Shaded Area


The probability that a z-score will be 1.54 or less = P(z ≤ 1.54)
= 0.93822 ≈ 0.9382

Second Approach: Center-Shaded Area


The probability that a z-score will be 1.54 or less
= P(z ≤ 1.54)
= 0.5 + 0.43822
= 0.93822 ≈ 0.9382

Third Approach: Texas Instruments (TI) technology
Please see the example: Texas Instruments (TI) calculators for Probability
The probability that a z-score will be 1.54 or less
= 0.9382198075 ≈ 0.9382

(c.) The graph that shows the probability that a z-score is 1.54 or more is option B.


(d.) First Approach: Left-Shaded Area
The probability that a z-score will be 1.54 or more
= P(z ≥ 1.54)
= 1 − 0.93822
= 0.06178 ≈ 0.0618

Second Approach: Center-Shaded Area


The probability that a z-score will be 1.54 or more
= P(z ≥ 1.54)
= 0.5 − 0.43822
= 0.06178 ≈ 0.0618

Third Approach: Texas Instruments (TI) technology
Please see the example: Texas Instruments (TI) calculators for Probability
The probability that a z-score will be 1.54 or more
= 0.0617801925 ≈ 0.0618

(e.) The graph that shows the probability that a z-score is between −1.5 and −1.05 is option A.


(f.) −1.5 = −1.5 (vertical); 0.00 (horizontal)
−1.05 = −1.0 (vertical); 0.05 (horizontal)

First Approach: Left-Shaded Area


The probability that a z-score is between −1.5 and −1.05
= P(−1.5 ≤ z ≤ −1.05)
= P(z ≤ −1.05) - P(z ≤ −1.5)
= 0.14686 − 0.06681
= 0.08005 ≈ 0.0801

Second Approach: Center-Shaded Area


The probability that a z-score is between −1.5 and −1.05
= P(−1.5 ≤ z ≤ −1.05)
= P(0 ≤ z ≤ −1.5) − = P(0 ≤ z ≤ −1.05)...symmetrical
= 0.43319 − 0.35314
= 0.08005 ≈ 0.0801

Third Approach: Texas Instruments (TI) technology
Please see the example: Texas Instruments (TI) calculators for Probability
The probability that a z-score is between −1.5 and −1.05
= 0.080018519 ≈ 0.0801

Let the shoe size that fits women with feet lengths of 21.6 cm through 25 cm = x
Let x₁ be the first shoe size
Let x₂ be the second shoe size

$ \mu = 22.8\;cm \\[3ex] \sigma = 1.2\;cm \\[3ex] x_1 = 21.6\;cm \\[3ex] x_2 = 25\;cm \\[3ex] $ 3rd Approach: Texas Instruments Calculators





$ P(21.6 \le x \le 25) = \\[3ex] = normalcdf(21.6, 25, 22.8, 1.2) \\[3ex] = 0.8079682921 \\[3ex] = 80.79682921\% \\[3ex] \approx 80.8\% \\[3ex] $ 1st Approach: Normal Distribution Table (Left Shaded Area)

$ z_1 = \dfrac{x_1 - \mu}{\sigma} \\[5ex] = \dfrac{21.6 - 22.8}{1.2} \\[5ex] = -1 \\[3ex] = -1.00 \\[3ex] P(z_1 \lt -1.00) = 0.15866 \\[5ex] z_2 = \dfrac{x_2 - \mu}{\sigma} \\[5ex] = \dfrac{25 - 22.8}{1.2} \\[5ex] = 1.833333333 \\[3ex] \approx 1.83 \\[3ex] P(z_2 \lt 1.83) = 0.96638 \\[5ex] P(21.6 \le x \le 25) = \\[3ex] = P(-1.00 \le z \le 1.83) = \\[3ex] = P(z \le 1.83) - P(z \le -1.00) \\[3ex] = 0.96638 - 0.15866 \\[3ex] = 0.80772 \\[3ex] = 80.772\% \\[3ex] \approx 80.8\% \\[3ex] $ 2nd Approach: Normal Distribution Table (Center Shaded Area)

$ P(21.6 \le x \le 25) = \\[3ex] = P(-1.00 \le z \le 1.83) = \\[3ex] = P(-1.00 \le z \le 0) + P(0 \le z \le 1.83) \\[3ex] = P(0 \le z \le 1) + P(0 \le z \le 1.83) \\[3ex] = 0.34134 + 0.46638 \\[3ex] = 0.80772 \\[3ex] = 80.772\% \\[3ex] \approx 80.8\% \\[3ex] $ About 80.8% of the women will be able to find shoes that fit in this store.

The percentage of boys aged 16 to 17 will not be able to find shoes that fit in this store are:
The ones whose shoe size is less than 24.6 cm and the ones whose shoe size is greater than 28.8 cm.
This is known as the Probability(away from)
Let x be the shoe size that is less than 24.6 cm or the shoe size that is greater than 28.8 cm
Let x₁ be the first shoe size
Let x₂ be the second shoe size

$ \mu = 25.4\;cm \\[3ex] \sigma = 1.5\;cm \\[3ex] x_1 = 24.6\;cm \\[3ex] x_2 = 28.8\;cm \\[3ex] $ 3rd Approach: Texas Instruments Calculators









$ P(x \le 24.6) \\[3ex] = normalcdf(-1000000, 24.6, 25.4, 1.5) \\[3ex] = 0.29690140546 \\[5ex] P(x \ge 28.8) \\[3ex] = normalcdf(28.8, 1000000, 25.4, 1.5) \\[3ex] = 0.0117052622 \\[5ex] P(away\;\;from) \\[3ex] = 0.29690140546 + 0.0117052622 \\[3ex] = 0.30860666766 \\[3ex] = 30.860666766\% \\[3ex] \approx 30.9\% \\[3ex] $ 1st Approach: Normal Distribution Table (Left Shaded Area)

$ z_1 = \dfrac{x_1 - \mu}{\sigma} \\[5ex] = \dfrac{24.6 - 25.4}{1.5} \\[5ex] = -0.533333333 \\[3ex] \approx -0.53 \\[3ex] P(z_1 \lt -0.53) = 0.29806 \\[5ex] z_2 = \dfrac{x_2 - \mu}{\sigma} \\[5ex] = \dfrac{28.8 - 25.4}{1.5} \\[5ex] = 2.266666667 \\[3ex] \approx 2.27 \\[3ex] P(z_2 \lt 2.27) = 0.98840 \\[5ex] P(z_2 \gt 2.27) \\[3ex] = 1 - 0.98840 \\[3ex] = 0.0116 \\[3ex] P(away\;\;from) \\[3ex] = P(z_1 \lt -0.53) + P(z_2 \gt 2.27) \\[3ex] = 0.29806 + 0.0116 \\[3ex] = 0.30966 \\[3ex] = 30.966\% \\[3ex] \approx 31.0\% \\[3ex] $ About 30.9% of the boys aged 16 to 17 will not be able to find shoes that fit in this store.

Let the birth weight of elephants at the 85th percentile = x
85th percentile = 85% lie below it = 0.85 lie below it

$ P(z \lt value) = 0.85 \\[3ex] Inverse\;\;Normal\;\;Distribution \implies \\[3ex] \underline{1st\;\;Approach:}\;\;\boldsymbol{Normal\;\;Distribution\;\;Table} \\[3ex] \underline{Left-shaded\;\;Area} \\[3ex] value = ? ...not\;\;found \\[5ex] \boldsymbol{Interpolate} \\[3ex] P(z \lt value\;1) = P(z \lt 1.03) = 0.84849 \\[3ex] P(z \lt value) = 0.85 \\[3ex] P(z \lt value\;2) = P(z \lt 1.04) = 0.85083 \\[3ex] \implies \\[3ex] \dfrac{z - 1.03}{1.04 - 1.03} = \dfrac{0.85 - 0.84849}{0.85083 - 0.84849} \\[5ex] \dfrac{z - 1.03}{0.01} = \dfrac{0.00151}{0.00234} \\[5ex] z - 1.03 = \dfrac{0.01 * 0.00151}{0.00234} \\[5ex] z - 1.03 = 0.0064529915 \\[3ex] z = 0.0064529915 + 1.03 \\[3ex] z = 1.036452991 \\[3ex] \mu = 240\;lbs \\[3ex] \sigma = 60\;lbs \\[3ex] x = \mu + z\sigma \\[3ex] x = 240 + 1.036452991(60) \\[3ex] x = 240 + 62.18717949 \\[3ex] x = 302.1871795 \\[3ex] x \approx 302\;lbs \\[5ex] $ 3rd Approach: Texas Instruments Calculators





$ P(z \lt value) = 0.85 \\[3ex] value = invNorm(0.85, 240, 60) \\[3ex] value = 302.1860028 \\[3ex] value \approx 302\lbs $

Empirical Rule: 68 - 95 - 99.7% Rule
Part of the Empirical Rule states that 95% of these babies have a birth weight within (below and above) 2 standard deviations from the mean

$ \mu = 3000\;grams \\[3ex] \sigma = ? \\[3ex] \mu - 2\sigma = 1600 \\[3ex] \mu - 1600 = 2\sigma \\[3ex] 2\sigma = \mu - 1600 \\[3ex] 2\sigma = 3000 - 1600 \\[3ex] 2\sigma = 1400 \\[3ex] \sigma = \dfrac{1400}{2} \\[5ex] \sigma = 700\;grams \\[3ex] \underline{Confirm\;\;the\;\;value\;\;of\;\;the\;\;standard\;\;deviation} \\[3ex] \mu + 2\sigma = 4400 \\[3ex] 2\sigma = 4400 - \mu \\[3ex] 2\sigma = 4400 - 3000 \\[3ex] 2\sigma = 1400 \\[3ex] \sigma = \dfrac{1400}{2} \\[5ex] \sigma = 700\;grams \\[3ex] \underline{For\;\;the\;\;baby} \\[3ex] x = 3497\;grams \\[3ex] z = \dfrac{x - \mu}{\sigma} \\[5ex] z = \dfrac{3497 - 3000}{700} \\[5ex] z = \dfrac{497}{700} \\[5ex] z = 0.71 $