2. a set of 120 test scores are normally distributed with a mean of 82 and a standard deviation of 5.
a) what percent of the scores are between 72 and 87?
b) what is the probability that a score is greater than 77?
c) what is the probability that a score is less than 82 or greater than 92?
d) about how many students scored outside two standard deviations of the mean?
3. the price of a gallon of regular gasoline at 75 gas stations across the state is normally distributed with a mean of $2.05 and a standard deviation of 4¢.
a) what percent of gas stations sell a gallon of regular gas for less than $1.97?
b) what percent of gas stations sell a gallon of regular gas for at least $2.17?
c) what is the probability that a gas station sells a gallon of regular gas for less than $1.97 or greater than $2.05?
d) about how many stations sell a gallon of regular gas for no more than $2.01?
4. mrs. fuller recently tested her 120 keyboarding students to see how many words per minute they can type. the results were normally distributed with a mean of 45 and a standard deviation of 6.
a) about how many students can type at least 39 words per minute?
b) about how many students can type within one standard deviation of the mean?
c) students need to be in the top 2% in order to be eligible for the national typing competition. if carla can type 56 wpm, is she eligible?

Question

2. a set of 120 test scores are normally distributed with a mean of 82 and a standard deviation of 5.
  a) what percent of the scores are between 72 and 87?
  b) what is the probability that a score is greater than 77?
  c) what is the probability that a score is less than 82 or greater than 92?
  d) about how many students scored outside two standard deviations of the mean?
3. the price of a gallon of regular gasoline at 75 gas stations across the state is normally distributed with a mean of $2.05 and a standard deviation of 4¢.
  a) what percent of gas stations sell a gallon of regular gas for less than $1.97?
  b) what percent of gas stations sell a gallon of regular gas for at least $2.17?
  c) what is the probability that a gas station sells a gallon of regular gas for less than $1.97 or greater than $2.05?
  d) about how many stations sell a gallon of regular gas for no more than $2.01?
4. mrs. fuller recently tested her 120 keyboarding students to see how many words per minute they can type. the results were normally distributed with a mean of 45 and a standard deviation of 6.
  a) about how many students can type at least 39 words per minute?
  b) about how many students can type within one standard deviation of the mean?
  c) students need to be in the top 2% in order to be eligible for the national typing competition. if carla can type 56 wpm, is she eligible?

Accepted Answer

### Problem 2 (Test Scores, Normal Distribution)
#### Part a)
# Explanation:
## Step1: Find z - scores
The mean $\mu = 82$, standard deviation $\sigma=5$. For $x = 72$, $z_1=\frac{72 - 82}{5}=\frac{- 10}{5}=- 2$. For $x = 87$, $z_2=\frac{87 - 82}{5}=\frac{5}{5}=1$.
## Step2: Use empirical rule
The empirical rule for normal distribution:
- Between $z=-2$ and $z = 0$: 47.5% (since between $\mu - 2\sigma$ and $\mu$ is 47.5%)
- Between $z = 0$ and $z = 1$: 34.1% (between $\mu$ and $\mu+\sigma$ is 34.1%)
- Total percentage between $z=-2$ and $z = 1$ is $47.5\%+34.1\% = 81.6\%$
# Answer:
$81.6\%$

#### Part b)
# Explanation:
## Step1: Find z - score for $x = 77$
$z=\frac{77 - 82}{5}=\frac{-5}{5}=-1$
## Step2: Use empirical rule
The area to the right of $z=-1$: The total area under the curve is 1 (or 100%). The area to the left of $z=-1$ is $15.9\%$ (since between $\mu - 2\sigma$ and $\mu-\sigma$ is 13.6%, and left of $\mu - 2\sigma$ is 2.3%, wait no, correct: The area to the left of $z=-1$ is $15.9\%$ (from empirical rule: left of $\mu-\sigma$ is $15.9\%$). So the area to the right (probability) is $100\% - 15.9\%=84.1\%$ (or 0.841)
# Answer:
$84.1\%$ (or 0.841)

#### Part c)
# Explanation:
## Step1: Analyze the two regions
- Probability less than 82: Since 82 is the mean, the area to the left of the mean in a normal distribution is 50% (or 0.5)
- Probability greater than 92: First find z - score for $x = 92$, $z=\frac{92 - 82}{5}=\frac{10}{5}=2$. The area to the right of $z = 2$ is 2.3% (from empirical rule, right of $\mu + 2\sigma$ is 2.3%)
## Step2: Add the two probabilities
$50\%+2.3\% = 52.3\%$ (or 0.523)
# Answer:
$52.3\%$ (or 0.523)

#### Part d)
# Explanation:
## Step1: Find percentage outside two standard deviations
From empirical rule, the percentage of data within two standard deviations of the mean is $95.4\%$, so the percentage outside is $100\% - 95.4\%=4.6\%$
## Step2: Calculate number of students
Number of students $=0.046	imes120 = 5.52\approx6$ (we round to nearest whole number)
# Answer:
About 6 students

### Problem 3 (Gasoline Prices, Normal Distribution)
#### Part a)
# Explanation:
## Step1: Find z - score for $x = 1.97$
Mean $\mu = 2.05$, standard deviation $\sigma = 0.04$ (assuming 4¢ is $0.04$). $z=\frac{1.97 - 2.05}{0.04}=\frac{-0.08}{0.04}=-2$
## Step2: Use empirical rule
The area to the left of $z=-2$: From empirical rule, the area left of $\mu - 2\sigma$ is $2.3\%$ (since between $\mu - 3\sigma$ and $\mu - 2\sigma$ is 2.2% and left of $\mu - 3\sigma$ is 0.1%, total $2.3\%$)
# Answer:
$2.3\%$

#### Part b)
# Explanation:
## Step1: Find z - score for $x = 2.17$
$z=\frac{2.17 - 2.05}{0.04}=\frac{0.12}{0.04}=3$
## Step2: Use empirical rule
The area to the right of $z = 3$ is 0.1% (from empirical rule, right of $\mu+3\sigma$ is 0.1%)
# Answer:
$0.1\%$

#### Part c)
# Explanation:
## Step1: Analyze the two regions
- Probability less than 1.97: As in part a, it's 2.3%
- Probability greater than 2.05: Since 2.05 is the mean, the area to the right of the mean is 50%
## Step2: Add the two probabilities
$2.3\%+50\% = 52.3\%$
# Answer:
$52.3\%$

#### Part d)
# Explanation:
## Step1: Find z - score for $x = 2.01$
$z=\frac{2.01 - 2.05}{0.04}=\frac{-0.04}{0.04}=-1$
## Step2: Use empirical rule
The area to the left of $z=-1$: From empirical rule, the area left of $\mu-\sigma$ is $15.9\%$? Wait no, wait: The area between $\mu - 2\sigma$ and $\mu-\sigma$ is 13.6%, left of $\mu - 2\sigma$ is 2.3%, so left of $\mu-\sigma$ (z=-1) is $13.6\%+2.3\%+0.1\%$? No, correct empirical rule:
- $\mu - 3\sigma$ to $\mu - 2\sigma$: 2.2%
- $\mu - 2\sigma$ to $\mu-\sigma$: 13.6%
- $\mu-\sigma$ to $\mu$: 34.1%
- $\mu$ to $\mu+\sigma$: 34.1%
- $\mu+\sigma$ to $\mu + 2\sigma$: 13.6%
- $\mu + 2\sigma$ to $\mu + 3\sigma$: 2.2%
- $\mu + 3\sigma$ and above: 0.1%
- $\mu - 3\sigma$ and below: 0.1%

So for $z=-1$ ( $\mu-\sigma$), the area to the left is $0.1\%+2.2\%+13.6\%=15.9\%$? Wait no, the total area to the left of $z=-1$ is $15.9\%$ (since the area from $-\infty$ to $\mu-\sigma$ is $15.9\%$). Wait, but let's recalculate. The area between $\mu - 3\sigma$ and $\mu$ is $0.1\%+2.2\%+13.6\%+34.1\% = 50\%$. So left of $\mu-\sigma$ (z=-1) is $0.1\%+2.2\%+13.6\%=15.9\%$? Wait, no, the area from $\mu - 3\sigma$ to $\mu-\sigma$ is $0.1\%+2.2\%+13.6\% = 15.9\%$, and left of $\mu - 3\sigma$ is 0.1%, so total left of $\mu-\sigma$ is $15.9\%+0.1\%$? No, I think I made a mistake. The correct empirical rule percentages (in order from left to right, starting from far left):

- Less than $\mu - 3\sigma$: 0.1%
- $\mu - 3\sigma$ to $\mu - 2\sigma$: 2.2%
- $\mu - 2\sigma$ to $\mu-\sigma$: 13.6%
- $\mu-\sigma$ to $\mu$: 34.1%
- $\mu$ to $\mu+\sigma$: 34.1%
- $\mu+\sigma$ to $\mu + 2\sigma$: 13.6%
- $\mu + 2\sigma$ to $\mu + 3\sigma$: 2.2%
- Greater than $\mu + 3\sigma$: 0.1%

So the area to the left of $\mu-\sigma$ (z=-1) is $0.1\%+2.2\%+13.6\% = 15.9\%$? Wait, no, the area from $-\infty$ to $\mu-\sigma$ is $0.1\%+2.2\%+13.6\%=15.9\%$? Wait, no, the area from $-\infty$ to $\mu$ is 50%, so the area from $-\infty$ to $\mu-\sigma$ is $50\% - 34.1\%=15.9\%$ (since from $\mu-\sigma$ to $\mu$ is 34.1%). Yes, that's correct. So the area to the left of $z=-1$ (x = 2.01) is 15.9%? Wait, but in the graph, it's marked as 2.3% for left of 1.97 (which is $\mu - 2\sigma$). Wait, maybe the standard deviation is 4¢, so $\mu=2.05$, $\sigma = 0.04$. Then:

- $\mu - 3\sigma=2.05 - 0.12 = 1.93$
- $\mu - 2\sigma=2.05 - 0.08 = 1.97$
- $\mu-\sigma=2.05 - 0.04 = 2.01$
- $\mu=2.05$
- $\mu+\sigma=2.05 + 0.04 = 2.09$
- $\mu + 2\sigma=2.05 + 0.08 = 2.13$
- $\mu + 3\sigma=2.05 + 0.12 = 2.17$

So for $x = 2.01$ (which is $\mu-\sigma$), the area to the left of $x = 2.01$ is the area less than $\mu-\sigma$. From the empirical rule, the area less than $\mu-\sigma$ is $0.1\%+2.2\%+13.6\%=15.9\%$? Wait, no, the area from $\mu - 3\sigma$ to $\mu - 2\sigma$ is 2.2%, $\mu - 2\sigma$ to $\mu-\sigma$ is 13.6%, and less than $\mu - 3\sigma$ is 0.1%. So total less than $\mu-\sigma$ is $0.1\%+2.2\%+13.6\% = 15.9\%$. But the number of gas stations is 75. So the number of stations with price no more than $2.01$ is $0.159	imes75\approx11.925\approx12$? Wait, but maybe I misread the graph. Wait the graph for problem 3 has 2.3% left of 1.97 (which is $\mu - 2\sigma$), which is $0.1\%+2.2\%=2.3\%$, which matches. Then for $\mu-\sigma$ (2.01), the area left of $\mu-\sigma$ is $2.3\%+13.6\%=15.9\%$? Wait, no, the area between $\mu - 2\sigma$ and $\mu-\sigma$ is 13.6%, so left of $\mu-\sigma$ is left of $\mu - 2\sigma$ (2.3%) plus between $\mu - 2\sigma$ and $\mu-\sigma$ (13.6%)? No, left of $\mu-\sigma$ is all the area from $-\infty$ to $\mu-\sigma$, which is left of $\mu - 3\sigma$ (0.1%) + between $\mu - 3\sigma$ and $\mu - 2\sigma$ (2.2%) + between $\mu - 2\sigma$ and $\mu-\sigma$ (13.6%) = 15.9%. So the number of stations is $0.159	imes75\approx12$ (rounded to nearest whole number)
## Step2: Calculate number of stations
Number of stations $=0.159	imes75\approx12$ (or using more accurate percentage: 15.9% of 75)
# Answer:
About 12 stations

### Problem 4 (Keyboarding Speeds, Normal Distribution)
#### Part a)
# Explanation:
## Step1: Find z - score for $x = 39$
Mean $\mu = 45$, standard deviation $\sigma = 6$. $z=\frac{39 - 45}{6}=\frac{-6}{6}=-1$
## Step2: Use empirical rule
The area to the right of $z=-1$: The area to the left of $z=-1$ is 15.9% (from empirical rule), so the area to the right is $100\% - 15.9\%=84.1\%$
## Step3: Calculate number of students
Number of students $=0.841	imes120\approx100.92\approx101$
# Answer:
About 101 students

#### Part b)
# Explanation:
## Step1: Percentage within one standard deviation
From empirical rule, the percentage of data within one standard deviation of the mean is $68.2\%$ (34.1% on each side of the mean)
## Step2: Calculate number of students
Number of students $=0.682	imes120\approx81.84\approx82$
# Answer:
About 82 students

#### Part c)
# Explanation:
## Step1: Find z - score for $x = 56$
$z=\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$ (wait, no, 56 - 45 = 11, 11/6≈1.83? Wait, no, 45+3*6=45 + 18=63, 45+2*6=57. So 56 is less than 57 ($\mu + 2\sigma$). The top 2% of data is above $z\approx2.05$ (from z - table, the area to the right of $z = 2.05$ is about 2%). Let's calculate the z - score for 56: $z=\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$. The area to the right of $z = 1.83$: From z - table, the area to the left of $z = 1.83$ is 0.9664, so the area to the right is $1 - 0.9664 = 0.0336$ or 3.36%, which is more than 2%. Wait, but maybe using empirical rule: The area above $\mu + 2\sigma$ (57) is 2.3% (from empirical rule, right of $\mu + 2\sigma$ is 2.3%). Since 56 is less than 57, the area above 56 is more than 2.3%? Wait, no, 57 is $\mu + 2\sigma=45 + 12 = 57$. So 56 is 1 less than 57, so z - score is $\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$. The top 2% corresponds to z - score of about 2.05 (since $P(Z>2.05)\approx0.02$). Since 1.83 < 2.05, the area above 56 is more than 2%, so Carla is not in the top 2%? Wait, no, wait: Wait, the mean is 45, standard deviation 6. So $\mu + 2\sigma=45 + 12 = 57$, $\mu + 3\sigma=45+18 = 63$. The area above $\mu + 2\sigma$ is 2.3% (from empirical rule: right of $\mu + 2\sigma$ is 2.2% + 0.1% = 2.3%). The area above $\mu + 1.83\sigma$: Let's use

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QUESTION IMAGE

Question

Explanation:

Problem 2 (Test Scores, Normal Distribution)

Part a)

Step1: Find z - scores

Step2: Use empirical rule

Step1: Find z - score for $x = 77$

Step2: Use empirical rule

Step1: Analyze the two regions

Step2: Add the two probabilities

Answer:

Part b)