The average age of chemical engineers is 37 years with a standard deviation of 4 years. If an engineering firm employs 25 chemical engineers, find the probability that the average age of the group is greater than 38.2 years old. If this is the case, would it be safe to assume that the chemical engineers in this group are generally much older than average.
Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury and the standard deviation is 5.6. Assume the variable is normally distributed:
A)If an individual is selected, find the probability that the individual;s pressure will be between 120 and 121.8
B) If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120 and 121.8
At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed:
A) If a proofreader from the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years.
B) If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders will be between 36 and 37.5 years
Z = (Sample mean-population mean)/SEm = (38.2-37)/SEm
SEm (Standard Error of the Mean) = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Use similar process for B sections of other questions
A) Z = (score-mean)/SD = (121.8-120)/5.6
Use the same table and similar process for both A questions.
The difference between A and B is that the former deals with a distribution of scores and the latter deals with a distribution of means.
Oh boy, let's have some fun with probabilities!
First, let's tackle the chemical engineers. We want to know the probability that the average age of the group is greater than 38.2 years old. Given that the average age is 37 with a standard deviation of 4, we can use the Central Limit Theorem to approximate this as a normal distribution.
Z = (38.2 - 37) / (4 / sqrt(25))
Z = 1.2 / (4 / 5)
Z = 1.2 / 0.8
Z = 1.5
Using a standard normal distribution table or calculator, we find that the probability corresponding to a Z-score of 1.5 is approximately 0.9332. So the probability that the average age of the group is greater than 38.2 years old is about 0.9332.
But does this mean the chemical engineers in this group are generally much older than average? Well, not necessarily! The probability only tells us about the likelihood of obtaining a sample mean of that size, given the population mean and standard deviation. It doesn't make any claims about the true age of the chemical engineers in the firm.
Now, let's dive into blood pressure. We're assuming the mean systolic blood pressure of normal adults is 120 with a standard deviation of 5.6.
A) If an individual is selected, the probability that their pressure will be between 120 and 121.8 is approximately 0.1708. So, there's about a 17.08% chance of falling within this range.
B) If a sample of 30 adults is randomly selected, we can still use the Central Limit Theorem to approximate the sample mean as a normal distribution. The standard deviation of the sample mean is the population standard deviation divided by the square root of the sample size: 5.6 / sqrt(30) ≈ 1.0223.
Now, we can calculate the Z-scores:
Z1 = (120 - 120) / 1.0223 ≈ 0
Z2 = (121.8 - 120) / 1.0223 ≈ 1.77
Using the standard normal distribution, the probability of the sample mean falling between 120 and 121.8 is approximately 0.9616. So, there's about a 96.16% chance of the sample mean being within this range.
Lastly, let's talk proofreaders. We're given a mean age of 36.2 with a standard deviation of 3.7.
A) If a proofreader is randomly selected, the probability that their age will be between 36 and 37.5 is approximately 0.3186. So, there's about a 31.86% chance of falling within this age range.
B) If a random sample of 15 proofreaders is selected, we can again use the Central Limit Theorem to approximate the sample mean as a normal distribution. The standard deviation of the sample mean is the population standard deviation divided by the square root of the sample size: 3.7 / sqrt(15) ≈ 0.9554.
Now, let's calculate the Z-scores:
Z1 = (36 - 36.2) / 0.9554 ≈ -0.209
Z2 = (37.5 - 36.2) / 0.9554 ≈ 1.371
Using the standard normal distribution, the probability of the sample mean falling between 36 and 37.5 is approximately 0.4146. So, there's about a 41.46% chance of the sample mean being within this range.
And there you have it! Probabilities can be quite amusing, don't you think?
For the first question, we can use the Central Limit Theorem since we have a sample size of 25. The distribution of the average age of chemical engineers will be approximately normal.
To find the probability that the average age is greater than 38.2 years old, we need to standardize the value using the z-score formula:
Z = (X - μ) / (σ / √n)
where:
X = 38.2
μ = 37 (average age)
σ = 4 (standard deviation)
n = 25 (sample size)
Calculating the z-score:
Z = (38.2 - 37) / (4 / √25)
Z = 1.2 / (4 / 5)
Z = 1.5
Now, we need to find the probability of the z-score being greater than 1.5 using a standard normal distribution table or calculator. The probability is approximately 0.0668.
Therefore, the probability that the average age of the group is greater than 38.2 years old is 0.0668.
However, we cannot assume that the chemical engineers in this group are generally much older than average based solely on this probability. The difference in average age is relatively small, and there could be other factors affecting the distribution of ages within the group.
Now let's move on to the second question about systolic blood pressure.
For part A) if an individual is selected, we need to find the probability that their pressure will be between 120 and 121.8 mmHg. Since the variable is normally distributed, we can use the z-score formula again.
Z = (X - μ) / σ,
where:
X = 121.8 (upper value)
μ = 120 (mean systolic blood pressure)
σ = 5.6 (standard deviation)
Calculating the z-scores for both values:
Z1 = (121.8 - 120) / 5.6
Z2 = (120 - 120) / 5.6
Now we can find the probability using a standard normal distribution table or calculator: P(Z1) - P(Z2).
For part B) if a sample of 30 adults is randomly selected, we also need to find the probability that the sample mean will be between 120 and 121.8 mmHg. Since the sample size is 30, we can still assume a normal distribution for the sample mean due to the Central Limit Theorem.
However, now we need to use the standard error of the mean, which is calculated as σ / √n:
SE = σ / √n,
where:
σ = 5.6 (standard deviation)
n = 30 (sample size)
Calculating the standard error:
SE = 5.6 / √30
Next, we need to calculate the z-scores for the sample mean using the formula:
Z1 = (121.8 - 120) / SE
Z2 = (120 - 120) / SE
Again, find the probability using a standard normal distribution table or calculator: P(Z1) - P(Z2).
Now let's move on to the last question about the mean age of proofreaders.
For part A) if a proofreader is randomly selected, we need to find the probability that their age will be between 36 and 37.5 years. Since the variable is normally distributed, we can use the z-score formula.
Z = (X - μ) / σ,
where:
X = 37.5 (upper value)
μ = 36.2 (mean age)
σ = 3.7 (standard deviation)
Calculating the z-scores:
Z1 = (37.5 - 36.2) / 3.7
Z2 = (36 - 36.2) / 3.7
Find the probability using a standard normal distribution table or calculator: P(Z1) - P(Z2).
For part B) if a random sample of 15 proofreaders is selected, we need to find the probability that the mean age of the proofreaders will be between 36 and 37.5 years. Since the sample size is 15, we can assume a normal distribution for the sample mean due to the Central Limit Theorem.
Calculate the standard error using the formula:
SE = σ / √n,
where:
σ = 3.7 (standard deviation)
n = 15 (sample size)
Calculating the standard error:
SE = 3.7 / √15
Next, calculate the z-scores for the sample mean using the formula:
Z1 = (37.5 - 36.2) / SE
Z2 = (36.0 - 36.2) / SE
Find the probability using a standard normal distribution table or calculator: P(Z1) - P(Z2).
To solve the given probability questions, we need to use the concept of the normal distribution. Before we proceed, note that we assume that the variables we are interested in (age, blood pressure, etc.) follow a normal distribution.
For each question, we will use the Z-score formula, which allows us to standardize our values and convert them into a standard normal distribution with a mean of 0 and a standard deviation of 1.
The Z-score formula is:
Z = (X - μ) / σ
Where:
Z is the standard score
X is the value of the random variable
μ is the mean of the distribution
σ is the standard deviation of the distribution
Now let's solve each question using this formula:
1) For the chemical engineers' average age:
The mean (μ) is given as 37 years, and the standard deviation (σ) is given as 4 years. Since we have a sample of 25 chemical engineers, the standard deviation of the sample mean (σ_m) will be σ / sqrt(n), where n is the sample size. Therefore, σ_m = 4 / sqrt(25) = 4 / 5 = 0.8.
To find the probability that the average age of the group is greater than 38.2 years, we need to calculate the Z-score for 38.2 using the formula mentioned above: Z = (38.2 - 37) / 0.8 = 1.5.
Now, we can find the probability using a Z-table or statistical software. The area under the normal curve to the right of Z = 1.5 represents the probability that the average age is greater than 38.2 years old.
2) For the individual's systolic blood pressure:
The mean (μ) is given as 120 mmHg, and the standard deviation (σ) is given as 5.6 mmHg. To find the probability that an individual's pressure will be between 120 and 121.8 mmHg, we need to calculate the Z-scores for these values:
Z1 = (120 - 120) / 5.6 = 0
Z2 = (121.8 - 120) / 5.6
After calculating the Z-scores, we can find the probability using a Z-table or statistical software. The area under the normal curve between Z1 and Z2 represents the probability that the individual's pressure will be between those values.
3) For the sample mean:
The mean (μ) and standard deviation (σ) remain the same as in the previous question. Since we now have a sample of 30 adults, the standard deviation of the sample mean (σ_m) will be σ / sqrt(n). Thus, σ_m = 5.6 / sqrt(30).
To find the probability that the sample mean will be between 120 and 121.8 mmHg, we need to calculate the Z-scores as before:
Z1 = (120 - 120) / (5.6 / sqrt(30))
Z2 = (121.8 - 120) / (5.6 / sqrt(30))
After calculating the Z-scores, we can find the probability using a Z-table or statistical software. The area under the normal curve between Z1 and Z2 represents the probability that the sample mean will be between those values.
4) For the proofreaders' individual ages:
The mean (μ) is given as 36.2 years, and the standard deviation (σ) is given as 3.7 years. To find the probability that a proofreader's age will be between 36 and 37.5 years, we need to calculate the Z-scores for these values:
Z1 = (36 - 36.2) / 3.7
Z2 = (37.5 - 36.2) / 3.7
After calculating the Z-scores, we can find the probability using a Z-table or statistical software. The area under the normal curve between Z1 and Z2 represents the probability that the proofreader's age will be between those values.
5) For the mean age of the Proofreaders:
The mean (μ) and standard deviation (σ) remain the same as in the previous question. Since we now have a sample of 15 proofreaders, the standard deviation of the sample mean (σ_m) will be σ / sqrt(n). Thus, σ_m = 3.7 / sqrt(15).
To find the probability that the mean age of the proofreaders will be between 36 and 37.5 years, we need to calculate the Z-scores as before:
Z1 = (36 - 36.2) / (3.7 / sqrt(15))
Z2 = (37.5 - 36.2) / (3.7 / sqrt(15))
After calculating the Z-scores, we can find the probability using a Z-table or statistical software. The area under the normal curve between Z1 and Z2 represents the probability that the mean age of the proofreaders will be between those values.
Note: Calculating exact probabilities using Z-tables can be time-consuming. Hence, it is recommended to use statistical software like Excel or R that provides built-in functions to calculate these probabilities more efficiently.