for questions 7 and 8, draw the normal distribution curve, then answer the questions.
7. a set of 125 golf scores are normally distributed with a mean of 76 and a standard deviation of 3.
a) what percent of the scores are between 67 and 85?
b) what is the probability that a score is no more than 79?
c) about how many scores fell between one standard deviation of the mean?
8. the talk - time battery life of a group of cell phones is normally distributed with a mean of 5 hours and a standard deviation of 15 minutes.
a) what percent of the phones have a battery life of at least 4 hours and 45 minutes?
b) what percent of the phones have a battery life between 4.5 hours and 5.25 hours?
c) what percent of the phones have a battery life less than 5 hours or greater than 5.5 hours?
9. the number of hours that the employees at the grocery store worked last week is normally distributed with a mean of 24 and a standard deviation of 6. if there are 60 total employees, approximately how many worked at least 30 hours last week?
10. the grade point average (gpa) of the students at lakeview high school is normally distributed with a mean of 3.1 and a standard deviation of 0.3. if there are 1800 students enrolled at the school, approximately how many have a gpa between 2.5 and 3.7?

Question

Accepted Answer

### Question 7a
# Explanation:
## Step1: Find how many standard deviations 67 and 85 are from the mean.
The mean ($\mu$) is 76, standard deviation ($\sigma$) is 3.
For 67: $z_1=\frac{67 - 76}{3}=\frac{-9}{3}=-3$
For 85: $z_2=\frac{85 - 76}{3}=\frac{9}{3}=3$

## Step2: Use the empirical rule for normal distribution.
The empirical rule states that for a normal distribution:
- Approximately 68% of data is within $\mu\pm\sigma$
- Approximately 95% of data is within $\mu\pm2\sigma$
- Approximately 99.7% of data is within $\mu\pm3\sigma$

Since 67 is $\mu - 3\sigma$ and 85 is $\mu + 3\sigma$, the percentage of data between them is approximately 99.7%.

# Answer:
99.7%

### Question 7b
# Explanation:
## Step1: Find the z - score for 79.
$\mu = 76$, $\sigma = 3$.
$z=\frac{79 - 76}{3}=\frac{3}{3}=1$

## Step2: Use the empirical rule.
The area to the left of $z = 1$ (since we want scores no more than 79, i.e., $\leq79$) can be found using the empirical rule. The area within $\mu-\sigma$ to $\mu+\sigma$ is 68%, so the area to the left of $\mu+\sigma$ (79) is $50\%+\frac{68\%}{2}=84\%$ (more precisely, using the empirical rule approximations, the area to the left of $z = 1$ is approximately 84.13%, but for simplicity with the empirical rule, we can say approximately 84% or 84.1%).

# Answer:
Approximately 84.1% (or 84% using basic empirical rule)

### Question 7c
# Explanation:
## Step1: Recall the empirical rule for 1 standard deviation.
The empirical rule states that approximately 68% of data in a normal distribution lies within 1 standard deviation of the mean ($\mu\pm\sigma$).

## Step2: Calculate the number of scores.
Total number of scores is 125.
Number of scores within 1 standard deviation $= 0.68	imes125 = 85$

# Answer:
85

### Question 8a
# Explanation:
## Step1: Convert 4 hours 45 minutes to hours.
45 minutes $=\frac{45}{60}=0.75$ hours, so 4 hours 45 minutes $= 4.75$ hours.
Mean ($\mu$) = 5 hours, standard deviation ($\sigma$) = 15 minutes = 0.25 hours.

## Step2: Find the z - score for 4.75 hours.
$z=\frac{4.75 - 5}{0.25}=\frac{-0.25}{0.25}=-1$

## Step3: Use the empirical rule.
We want the percentage of phones with battery life at least 4.75 hours (i.e., $x\geq4.75$). The area to the right of $z=-1$:
The area within $\mu-\sigma$ to $\mu+\sigma$ is 68%, so the area to the right of $\mu-\sigma$ (4.75) is $50\%+\frac{68\%}{2}=84.1\%$ (since the normal distribution is symmetric, the area to the left of $z = - 1$ is $50\%-\frac{68\%}{2}=16\%$, so the area to the right is $100\% - 16\%=84\%$ or more precisely 84.1%).

# Answer:
84.1%

### Question 8b
# Explanation:
## Step1: Convert 4.5 hours and 5.25 hours to z - scores.
$\mu = 5$ hours, $\sigma = 0.25$ hours.
For 4.5 hours: $z_1=\frac{4.5 - 5}{0.25}=\frac{-0.5}{0.25}=-2$
For 5.25 hours: $z_2=\frac{5.25 - 5}{0.25}=\frac{0.25}{0.25}=1$

## Step2: Use the empirical rule.
- Area within $\mu - 2\sigma$ to $\mu$: 47.5% (since 95% is within $\mu\pm2\sigma$, so half is 47.5%)
- Area within $\mu$ to $\mu+\sigma$: 34.1% (since 68% is within $\mu\pm\sigma$, so half is 34.1%)

Total area between 4.5 ($\mu - 2\sigma$) and 5.25 ($\mu+\sigma$) is $47.5\%+34.1\% = 81.6\%$

# Answer:
81.6%

### Question 8c
# Explanation:
## Step1: Find z - scores for 5 hours and 5.5 hours.
$\mu = 5$ hours, $\sigma = 0.25$ hours.
For 5 hours: $z_1=\frac{5 - 5}{0.25}=0$
For 5.5 hours: $z_2=\frac{5.5 - 5}{0.25}=\frac{0.5}{0.25}=2$

## Step2: Find the areas.
- Area less than 5 hours (z = 0): 50% (since the mean is at z = 0, half the data is below)
- Area greater than 5.5 hours (z = 2): The area to the right of z = 2. Since 95% is within $\mu\pm2\sigma$, the area outside is 5%, so the area to the right of z = 2 is $\frac{5\%}{2}=2.5\%$

## Step3: Total area.
Total percentage $=50\%+2.5\% = 52.5\%$

# Answer:
52.5%

### Question 9
# Explanation:
## Step1: Find the z - score for 30 hours.
$\mu = 24$, $\sigma = 6$.
$z=\frac{30 - 24}{6}=\frac{6}{6}=1$

## Step2: Find the area to the right of z = 1.
Using the empirical rule, the area within $\mu-\sigma$ to $\mu+\sigma$ is 68%, so the area to the right of $\mu+\sigma$ (z = 1) is $\frac{100\% - 68\%}{2}=16\%$ (more precisely, 15.87%).

## Step3: Calculate the number of employees.
Total employees = 60.
Number of employees who worked at least 30 hours $=0.16	imes60 = 9.6\approx10$ (or using 15.87%: $0.1587	imes60\approx9.52\approx10$)

# Answer:
Approximately 10

### Question 10
# Explanation:
## Step1: Find z - scores for 2.5 and 3.7.
$\mu = 3.1$, $\sigma = 0.3$.
For 2.5: $z_1=\frac{2.5 - 3.1}{0.3}=\frac{-0.6}{0.3}=-2$
For 3.7: $z_2=\frac{3.7 - 3.1}{0.3}=\frac{0.6}{0.3}=2$

## Step2: Use the empirical rule.
The area within $\mu - 2\sigma$ to $\mu+2\sigma$ is approximately 95% (since 95% of data in a normal distribution is within 2 standard deviations of the mean).

## Step3: Calculate the number of students.
Total students = 1800.
Number of students with GPA between 2.5 and 3.7 $=0.95	imes1800 = 1710$

# Answer:
1710

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QUESTION IMAGE

Question

Explanation:

Question 7a

Step1: Find how many standard deviations 67 and 85 are from the mean.

Step2: Use the empirical rule for normal distribution.

Step1: Find the z - score for 79.

Step2: Use the empirical rule.

Step1: Recall the empirical rule for 1 standard deviation.

Step2: Calculate the number of scores.

Answer:

Question 7b