Assume the weights of men are normally distributed having a mean of 185 lbs and a standard deviation of 17 pounds.
A. Calculate the probability a randomly selected man weights less than 225 lbs
B. Approximately 72% of men’s weights exceed how many pounds.
C. In a random example of 50 men, calculate the probability their mean weights is no more than 180 pounds.
D. In a random example of 50 men, approximately 8% of all possible men’s weights will exceed how many pounds.
A, B. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (B use.72) related to the Z score.
Last two problems relate to distribution of means rather than a distribution of raw scores.
C, D. Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
A. To calculate the probability that a randomly selected man weighs less than 225 lbs, we can find the z-score corresponding to this weight and then use the standard normal distribution table.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = 225 lbs (desired weight)
μ = 185 lbs (mean weight)
σ = 17 lbs (standard deviation)
z = (225 - 185) / 17
z = 40 / 17
z ≈ 2.35
Now, we can use the standard normal distribution table or a calculator to find the probability corresponding to a z-score of 2.35. From the table or calculator, we find that the probability is approximately 0.9904.
Therefore, the probability that a randomly selected man weighs less than 225 lbs is approximately 0.9904 or 99.04%.
B. To find the weight that approximately 72% of men exceed, we need to find the z-score corresponding to this percentile.
Using the standard normal distribution table or a calculator, we find the z-score corresponding to a percentile of 72% is approximately 0.68.
Now, we can use the z-score formula to find the weight:
z = (x - μ) / σ
Where:
z = 0.68 (z-score)
μ = 185 lbs (mean weight)
σ = 17 lbs (standard deviation)
0.68 = (x - 185) / 17
Simplifying the equation:
0.68 * 17 = x - 185
11.56 = x - 185
x = 11.56 + 185
x ≈ 196.56
Approximately 72% of men's weights exceed 196.56 pounds.
C. To calculate the probability that the mean weight of a random sample of 50 men is no more than 180 pounds, we will use the sampling distribution of the sample mean.
The mean of the sampling distribution of the sample mean is the same as the population mean, μ = 185 lbs.
The standard deviation of the sampling distribution of the sample mean, also known as the standard error, can be calculated using the formula:
σ / √n
Where:
σ = 17 lbs (standard deviation of the population)
n = 50 (sample size)
Standard error = 17 / √50
Now, we can calculate the z-score:
z = (x - μ) / standard error
x = 180 lbs (desired mean weight)
μ = 185 lbs (population mean weight)
standard error = 17 / √50
z = (180 - 185) / (17 / √50)
z = -5 / (17 / √50)
z ≈ -1.18
Using the standard normal distribution table or a calculator, we find that the probability corresponding to a z-score of -1.18 is approximately 0.1190.
Therefore, the probability that the mean weight of a random sample of 50 men is no more than 180 pounds is approximately 0.1190 or 11.90%.
D. To find the weight that approximately 8% of all possible men's weights will exceed in a random sample of 50 men, we need to find the z-score corresponding to this percentile.
Using the standard normal distribution table or a calculator, we find the z-score corresponding to a percentile of 92% (since 100% - 8% = 92%) is approximately 1.41.
Now, we can use the z-score formula to find the weight:
z = (x - μ) / σ
Where:
z = 1.41 (z-score)
μ = 185 lbs (mean weight)
σ = 17 lbs (standard deviation)
1.41 = (x - 185) / 17
Simplifying the equation:
1.41 * 17 = x - 185
23.97 = x - 185
x = 23.97 + 185
x ≈ 208.97
Approximately 8% of all possible men's weights will exceed 208.97 pounds.
A. To calculate the probability that a randomly selected man weighs less than 225 lbs, we can use the standard normal distribution table. First, we need to standardize the value 225 using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
z = (225 - 185) / 17
z = 40 / 17
z ≈ 2.35
Looking up the z-score of 2.35 in the standard normal distribution table, we find that the corresponding probability is approximately 0.9904. Therefore, the probability that a randomly selected man weighs less than 225 lbs is approximately 0.9904 or 99.04%.
B. To determine the weight that approximately 72% of men's weights exceed, we need to find the z-score that corresponds to a percentile of 72%. We can do this by looking up the z-score in the standard normal distribution table.
A percentile of 72% corresponds to a z-score of approximately 0.65. Now, we can use the formula z = (x - μ) / σ and solve for x by rearranging the formula:
0.65 = (x - 185) / 17
0.65 * 17 = x - 185
11.05 + 185 = x
x ≈ 196.05
Therefore, approximately 72% of men's weights exceed 196.05 pounds.
C. To calculate the probability that the mean weight of a random sample of 50 men is no more than 180 pounds, we need to calculate the z-score for this value. Since we are dealing with sample means, we will use the standard error of the mean (SEM) instead of the standard deviation.
SEM = σ / sqrt(n)
SEM = 17 / sqrt(50)
SEM ≈ 2.407
Now we can calculate the z-score using the formula z = (x - μ) / SEM:
z = (180 - 185) / 2.407
z ≈ -2.07
Looking up a z-score of -2.07 in the standard normal distribution table, we find that the corresponding probability is approximately 0.0192. Therefore, the probability that the mean weight of a random sample of 50 men is no more than 180 pounds is approximately 0.0192 or 1.92%.
D. To determine the weight that approximately 8% of all possible men's weights will exceed in a random sample of 50 men, we need to calculate the z-score for this percentile. Using the standard normal distribution table, we can find the z-score that corresponds to the 92nd percentile (100% - 8%).
A percentile of 92% corresponds to a z-score of approximately 1.405. Now, we can use the formula z = (x - μ) / σ and solve for x by rearranging the formula:
1.405 = (x - 185) / 17
1.405 * 17 = x - 185
23.85 + 185 = x
x ≈ 208.85
Therefore, approximately 8% of all possible men's weights will exceed 208.85 pounds in a random sample of 50 men.