Solved NORMAL DISTRIBUTION WORKSHEET Name Section | Chegg.com - Free Printable

Here are the step-by-step solutions for the problems in the image. These problems rely on the Empirical Rule (also known as the 68-95-99.7 rule) for Normal Distributions.

Key Concepts to Remember:
* Mean ($\mu$): The center of the bell curve.
* Standard Deviation ($\sigma$): The distance from the mean to the inflection point.
* The Empirical Rule:
* 68% of data falls within 1 standard deviation of the mean.
* 95% of data falls within 2 standard deviations of the mean.
* 99.7% of data falls within 3 standard deviations of the mean.
* Because the curve is symmetric, 50% of the data is above the mean and 50% is below.

---

Problem 21

Given: The shaded area represents approximately 95% of the scores. The range shown is from 78 to 92.

a) What is the mean?
According to the Empirical Rule, 95% of the data lies within 2 standard deviations of the mean. This means the interval from 78 to 92 represents $\mu - 2\sigma$ to $\mu + 2\sigma$.
The mean is exactly in the middle of this range.
$$ \text{Mean} = \frac{78 + 92}{2} = \frac{170}{2} = 85 $$

b) What is the standard deviation?
The distance from the mean (85) to the upper end of the 95% interval (92) represents 2 standard deviations ($2\sigma$).
$$ 2\sigma = 92 - 85 = 7 $$
$$ \sigma = \frac{7}{2} = 3.5 $$

---

Problem 22

Given: Mean ($\mu$) = 106, Standard Deviation ($\sigma$) = 2.

a) Draw and label the Normal curve.
* Center: 106
* Right side marks: $106+2=108$, $108+2=110$, $110+2=112$
* Left side marks: $106-2=104$, $104-2=102$, $102-2=100$

b) What percentage of boxes contain more than 104 bolts?
104 is one standard deviation below the mean ($106 - 2$).
We know that 50% of the data is above the mean (106).
Between the mean and 1 SD below (104 to 106), there is half of 68%, which is 34%.
Total percentage > 104 = (Area from 104 to 106) + (Area above 106)
$$ 34\% + 50\% = 84\% $$

c) What percentage of boxes contain more than 110 bolts?
110 is two standard deviations above the mean ($106 + 2 + 2$).
95% of the data is within 2 SDs (between 102 and 110).
The remaining data outside this range is $100\% - 95\% = 5\%$.
Since the curve is symmetric, half of that 5% is above 110.
$$ \frac{5\%}{2} = 2.5\% $$

d) What percentage of boxes contain less than 108 bolts?
108 is one standard deviation above the mean.
50% of the data is below the mean.
Between the mean and 1 SD above (106 to 108), there is half of 68%, which is 34%.
Total percentage < 108 = (Area below 106) + (Area from 106 to 108)
$$ 50\% + 34\% = 84\% $$

e) What percentage of boxes contain less than 100 bolts?
100 is three standard deviations below the mean ($106 - 2 - 2 - 2$).
99.7% of the data is within 3 SDs (between 100 and 112).
The remaining data outside this range is $100\% - 99.7\% = 0.3\%$.
Half of that is below 100.
$$ \frac{0.3\%}{2} = 0.15\% $$

f) What percentage of boxes contain between 102 and 112 bolts?
102 is 2 SDs below the mean. 112 is 3 SDs above the mean.
* Area from mean to 2 SDs below: Half of 95% = 47.5%
* Area from mean to 3 SDs above: Half of 99.7% = 49.85%
* Total = $47.5\% + 49.85\% = 97.35\%$

g) What percentage of boxes contain between 100 and 106 bolts?
100 is 3 SDs below the mean. 106 is the mean.
This represents exactly half of the middle 99.7%.
$$ \frac{99.7\%}{2} = 49.85\% $$

---

Problem 23

Given: Mean ($\mu$) = 40 inches, Standard Deviation ($\sigma$) = 9 inches.

a) Draw and label the Normal curve.
* Center: 40
* Right side marks: $40+9=49$, $49+9=58$, $58+9=67$
* Left side marks: $40-9=31$, $31-9=22$, $22-9=13$

b) 97.5% of Ixxians are over \_\_\_\_\_ inches tall?
If 97.5% are *over* a certain height, then only 2.5% are *under* that height.
Looking at the left tail: 2.5% corresponds to the area below 2 standard deviations from the mean.
$$ \text{Mean} - 2\sigma = 40 - 2(9) = 40 - 18 = 22 $$

c) 16% of Ixxians are over \_\_\_\_\_ inches tall?
If 16% are *over*, then 84% are *under*.
84% corresponds to the area up to 1 standard deviation above the mean (50% below mean + 34% between mean and +1$\sigma$).
$$ \text{Mean} + 1\sigma = 40 + 9 = 49 $$

d) 50% of Ixxians are under \_\_\_\_\_ inches tall?
In a normal distribution, 50% of the data is always below the mean.
$$ \text{Mean} = 40 $$

e) 97.5% of Ixxians are under \_\_\_\_\_ inches tall?
This leaves 2.5% in the upper tail.
2.5% in the upper tail corresponds to 2 standard deviations above the mean.
$$ \text{Mean} + 2\sigma = 40 + 2(9) = 40 + 18 = 58 $$

f) The most "average" 68% of Ixxians are between \_\_\_\_\_ and \_\_\_\_\_ inches tall?
The middle 68% is always within 1 standard deviation of the mean.
Lower bound: $40 - 9 = 31$
Upper bound: $40 + 9 = 49$

g) 84% of Ixxians are over \_\_\_\_\_ inches tall?
If 84% are *over*, then 16% are *under*.
16% in the lower tail corresponds to being 1 standard deviation below the mean (since 50% - 34% = 16%).
$$ \text{Mean} - 1\sigma = 40 - 9 = 31 $$

──────────────────────────────────────

Final Answer:

21.
a) 85
b) 3.5

22.
a) [Curve centered at 106 with marks at 100, 102, 104, 108, 110, 112]
b) 84%
c) 2.5%
d) 84%
e) 0.15%
f) 97.35%
g) 49.85%

23.
a) [Curve centered at 40 with marks at 13, 22, 31, 49, 58, 67]
b) 22
c) 49
d) 40
e) 58
f) 31 and 49
g) 31