10 the number of hours in the lifespan of a certain brand of light bulb is normally distributed with a mean of 2387 hours and a standard deviation of 238 hours. to the nearest tenth of a percent, what percent of light bulbs have a lifespan of greater than 2750 hours?
1) 6.4%
2) 15.9%
3) 43.6%
4) 93.6%
11 the heights of the students at central high school can be modeled by a normal distribution with a mean of 68.1 and a standard deviation of 3.4 inches. according to this model, approximately what percent of the students would have a height less than 60 inches or greater than 75 inches?
1) 0.86%
2) 1.26%
3) 2.12%
4) 2.98%
12 the weight of a bag of pears at the local market averages 8 pounds with a standard deviation of 0.5 pound. the weights of all the bags of pears at the market closely follow a normal distribution. determine what percentage of bags, to the nearest integer, weighed less than 8.25 pounds.
13 the monthly unemployment rate of towns in the united states is approximately normally distributed with a mean rate of 5.2% and a standard deviation of 1.6%. determine the percentage of towns, to the nearest integer, that have a monthly unemployment rate greater than 6%.
14 the lifespan of a 60 - watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard deviation of 8.5 hours. if a 60 - watt lightbulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours?
1) 0.3803
2) 0.4612
3) 0.8415
4) 0.9612
15 the weights of bags of graseks chocolate candies are normally distributed with a mean of 4.3 ounces and a standard deviation of 0.05 ounces. what is the probability that a bag of these chocolate candies weighs less than 4.27 ounces?
1) 0.2257
2) 0.2743
3) 0.7257
4) 0.7757
16 two versions of a standardized test are given, an april version and a may version. the statistics for the april version show a mean score of 480 and a standard deviation of 24. the statistics for the may version show a mean score of 510 and a standard deviation of 20. assume the scores are normally distributed. joanne took the april version and scored in the interval 510 - 540. what is the probability, to the nearest ten thousandth, that a test paper selected at random from the april version scored in the same interval? maria took the may version. in what interval must maria score to claim she scored as well as joanne?
17 in 2013, approximately 1.6 million students took the critical reading portion of the sat exam. the mean score, the modal score, and the standard deviation were calculated to be 496, 430, and 115, respectively. which interval reflects 95% of the critical reading scores?
1) 430 ± 115
2) 430 ± 230
3) 496 ± 115
4) 496 ± 230
18 in a small city, there are 22 gas stations. the mean price for a gallon of regular gas was $2.12 with a standard deviation of $0.05. the distribution of the data was approximately normal. given this information, the middle 95% of the gas stations in this small city likely charge
1) $1.90 to $2.34 for a gallon of gas
2) $1.97 to $2.27 for a gallon of gas
3) $2.02 to $2.22 for a gallon of gas
4) $2.07 to $2.17 for a gallon of gas

Question

Accepted Answer

### Problem 10
# Explanation:
## Step 1: Calculate the z-score
The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x = 2750$, $\mu=2387$, and $\sigma = 238$.
$z=\frac{2750 - 2387}{238}=\frac{363}{238}\approx1.525$

## Step 2: Find the area to the right of the z - score
We use the standard normal distribution table (or a calculator with normal distribution functions). The area to the left of $z = 1.525$ is approximately $0.9365$. So the area to the right (the percentage of light bulbs with lifespan greater than 2750 hours) is $1 - 0.9365=0.0635\approx6.4\%$

# Answer: 1) 6.4%

### Problem 11
# Explanation:
## Step 1: Calculate z - scores for 60 and 75
For $x = 60$, $\mu = 68.1$, $\sigma=3.4$:
$z_1=\frac{60 - 68.1}{3.4}=\frac{- 8.1}{3.4}\approx - 2.38$
For $x = 75$, $\mu = 68.1$, $\sigma = 3.4$:
$z_2=\frac{75 - 68.1}{3.4}=\frac{6.9}{3.4}\approx2.03$

## Step 2: Find the areas
The area to the left of $z=-2.38$ is approximately $0.0087$. The area to the left of $z = 2.03$ is approximately $0.9788$. The area to the right of $z = 2.03$ is $1 - 0.9788 = 0.0212$. The total area (percentage of students with height less than 60 or greater than 75) is $0.0087+0.0212 = 0.0299\approx2.98\%$

# Answer: 4) 2.98%

### Problem 12
# Explanation:
## Step 1: Calculate the z - score
For $x = 8.25$, $\mu = 8$, $\sigma=0.5$:
$z=\frac{8.25 - 8}{0.5}=\frac{0.25}{0.5}=0.5$

## Step 2: Find the area to the left of the z - score
Using the standard normal distribution table, the area to the left of $z = 0.5$ is $0.6915$. To the nearest integer, this is $69\%$

# Answer: 69%

### Problem 13
# Explanation:
## Step 1: Calculate the z - score
For $x = 6\%$, $\mu = 5.2\%$, $\sigma=1.6\%$:
$z=\frac{6 - 5.2}{1.6}=\frac{0.8}{1.6}=0.5$

## Step 2: Find the area to the right of the z - score
The area to the left of $z = 0.5$ is $0.6915$. The area to the right is $1 - 0.6915 = 0.3085\approx31\%$

# Answer: 31%

### Problem 14
# Explanation:
## Step 1: Calculate z - scores for 1440 and 1465
For $x = 1440$, $\mu = 1450$, $\sigma=8.5$:
$z_1=\frac{1440 - 1450}{8.5}=\frac{- 10}{8.5}\approx - 1.18$
For $x = 1465$, $\mu = 1450$, $\sigma=8.5$:
$z_2=\frac{1465 - 1450}{8.5}=\frac{15}{8.5}\approx1.76$

## Step 2: Find the areas and subtract
The area to the left of $z=-1.18$ is approximately $0.1190$. The area to the left of $z = 1.76$ is approximately $0.9608$. The area between them is $0.9608 - 0.1190=0.8418\approx0.8415$

# Answer: 3) 0.8415

### Problem 15
# Explanation:
## Step 1: Calculate the z - score
For $x = 4.27$, $\mu = 4.3$, $\sigma=0.05$:
$z=\frac{4.27 - 4.3}{0.05}=\frac{- 0.03}{0.05}=- 0.6$

## Step 2: Find the area to the left of the z - score
Using the standard normal distribution table, the area to the left of $z=-0.6$ is $0.2743$

# Answer: 2) 0.2743

### Problem 16
#### Part 1: Probability for Joanne's interval (April version)
# Explanation:
## Step 1: Calculate z - scores for 510 and 540 (April: $\mu = 480$, $\sigma=24$)
For $x = 510$: $z_1=\frac{510 - 480}{24}=\frac{30}{24}=1.25$
For $x = 540$: $z_2=\frac{540 - 480}{24}=\frac{60}{24}=2.5$

## Step 2: Find the areas and subtract
The area to the left of $z = 1.25$ is $0.8944$. The area to the left of $z = 2.5$ is $0.9938$. The area between them is $0.9938 - 0.8944 = 0.0994\approx0.0994$ (to the nearest ten - thousandth)

#### Part 2: Maria's interval (May version: $\mu = 510$, $\sigma=20$)
We know that the z - scores for Joanne's interval are $z = 1.25$ and $z = 2.5$. We use the formula $x=\mu+z\sigma$
For $z = 1.25$: $x_1=510+1.25	imes20=510 + 25 = 535$
For $z = 2.5$: $x_2=510+2.5	imes20=510 + 50 = 560$
So Maria must score in the interval $535 - 560$

# Answer:
Probability: $0.0994$; Maria's interval: $535 - 560$

### Problem 17
# Explanation:
For a normal distribution, approximately 95% of the data lies within $2$ standard deviations of the mean. The mean is $496$ and the standard deviation is $115$. So the interval is $496\pm2	imes115=496\pm230$

# Answer: 4) $496\pm230$

### Problem 18
# Explanation:
For a normal distribution, approximately 95% of the data lies within $2$ standard deviations of the mean. The mean $\mu = 2.12$ and standard deviation $\sigma=0.05$. The interval is $\mu - 2\sigma$ to $\mu + 2\sigma$
$\mu-2\sigma=2.12-2	imes0.05=2.12 - 0.10 = 2.02$
$\mu + 2\sigma=2.12+2	imes0.05=2.12 + 0.10 = 2.22$
So the interval is $\$2.02$ to $\$2.22$

# Answer: 3) $\$2.02$ to $\$2.22$ for a gallon of gas

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 10

Step 1: Calculate the z-score

Step 2: Find the area to the right of the z - score

Step 1: Calculate z - scores for 60 and 75

Step 2: Find the areas

Step 1: Calculate the z - score

Step 2: Find the area to the left of the z - score

Answer:

Problem 11