How would I go about solving these problems?
In Exercise 9 we suggested the model N(1152,84) for weights in pounds of yearling Angus steers. What weight would you consider to be unusually low for such an animal? Explain.
If you are given the model N(1152,84),what are the cutoff values for the middle 40% of the weights?
I assume that you are talking about a normal distribution with a mean of 1152 and standard deviation of 84.
A Z score is the score in terms of standard deviations (SD), with -2 SD usually being considered low.
Z = (x - ƒÊ)/SD, where x = weight value and ƒÊ = mean. Solve for the unknown, once you find the Z score
The middle 40% is ƒÊ �} 20%.
Look up the various Z score values in the table in the back of your statistics booklabeled something like "areas under the normal distribution."
I hope this helps. Thanks for asking.
Well, well, well. Looks like we've got some math problems here. Don't worry, I'm here to make things a little more entertaining!
Now, when it comes to determining what weight would be considered unusually low for a yearling Angus steer, we need to take a look at the model N(1152,84). The "N" stands for normal distribution, which means we're dealing with a bell curve.
To find an unusually low weight, we can use something called z-scores. A z-score measures how far away a particular value is from the mean (in this case, 1152) in terms of standard deviations (in this case, 84).
Now, for z-scores, the general rule of thumb is that anything beyond -2 or +2 is considered unusual. So, if a weight falls more than 2 standard deviations below the mean, we can consider it unusually low for a yearling Angus steer.
Moving on to the next part of your question. You want to know the cutoff values for the middle 40% of the weights given the N(1152,84) model. Well, here's a little trick: the middle 40% is equivalent to the range between the 30th percentile and the 70th percentile.
So, what you need to do is find the z-scores that correspond to the 30th and 70th percentiles. Once you have those z-scores, you can multiply them by the standard deviation (84) and add/subtract the result from the mean (1152) to get your cutoff values.
I hope that brought a smile to your face while tackling these math problems! Remember, no need to stress, just keep clowning around!
To solve the first problem, you need to understand the concept of unusual or outlier values in a normal distribution. In a normal distribution, values that are located more than 1.5 times the interquartile range (IQR) below the first quartile (Q1) or above the third quartile (Q3) are considered to be outliers.
To determine the cutoff values for the middle 40% of the weights, you need to find the z-scores that correspond to the 30th and 70th percentiles of a standard normal distribution. The middle 40% of the distribution will fall between these two z-scores.
Here are the step-by-step solutions for both problems:
1. Determining an unusually low weight:
Step 1: Calculate the z-score using the formula:
z = (x - mean) / standard deviation
In this case, x = weight, mean = 1152, and standard deviation = 84.
Step 2: Calculate the z-score for a weight that is unusually low, by substituting the given values into the formula.
2. Determining the cutoff values for the middle 40%:
Step 1: Find the z-scores corresponding to the 30th and 70th percentiles (P30 and P70) of the standard normal distribution. These percentiles can be found using a standard normal distribution table or a statistical calculator.
Step 2: Calculate the actual values (weights) at these z-scores by using the formula:
x = z * standard deviation + mean
In this case, substitute the z-scores obtained in Step 1, the standard deviation value of 84, and the mean value of 1152 into the formula to calculate the cutoff values for the middle 40% of the weights.
Following these step-by-step instructions should help you solve the given problems.
To solve the first problem, determining an unusually low weight for yearling Angus steers, we can use statistical methods. In this case, we are given a normal distribution model with a mean (or average) weight of 1152 pounds and a standard deviation of 84 pounds.
To find the unusually low weight, we need to consider how different weights deviate from the mean. In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
To determine an unusually low weight, we can use the concept of z-scores. A z-score measures how many standard deviations a data point is away from the mean. Generally, values that are more than 2 or 3 standard deviations away from the mean are considered unusual.
To calculate the z-score for an unusually low weight, we can use the formula:
z = (x - mean) / standard deviation
where x is the weight we want to analyze. In this case, x would be the unusually low weight we want to find.
For example, let's say we consider any weight that falls below two standard deviations (or z < -2) to be unusually low:
z = (x - 1152) / 84
Using z = -2 and rearranging the formula, we can solve for the weight (x):
-2 = (x - 1152) / 84
-168 = x - 1152
x = 984
So, any weight below 984 pounds would be considered unusually low for a yearling Angus steer based on this model.
Moving on to the second problem, to find the cutoff values for the middle 40% of the weights, we can use z-scores. Since we know that roughly 40% of the data falls within one standard deviation on either side of the mean, we need to find the z-score that corresponds to the 30th percentile (10% on each tail).
To find this z-score, we can use a standard normal distribution table or a statistical software. For a standard normal distribution table, the closest z-score to the 30th percentile is approximately -0.524.
With this z-score, we can now calculate the cutoff values for the middle 40% of weights using the formula:
cutoff value = mean + (z * standard deviation)
lower cutoff value = 1152 + (-0.524 * 84)
lower cutoff value ≈ 1099.92 pounds
upper cutoff value = 1152 + (0.524 * 84)
upper cutoff value ≈ 1204.08 pounds
Therefore, the cutoff values for the middle 40% of weights would be approximately 1099.92 pounds and 1204.08 pounds based on the given normal distribution model N(1152,84).