Please help, need advise. Thanks
A
The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine? ì ó
B
The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. Approximately 83% of the vitamins will have at least how many grams of pyridoxine? ì ó
I believe you copied these questions incorrectly. The first one only makes sense if you meant:
The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with (mean) = 110 grams and (standard deviation) = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine?
You also left out the words "mean" and "standard deviation" in the second question.
Both questions can be answered with normal distribution tables or computational tools.
For a time-saving computational tool for problems like this, see
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
I get 34.5% for your first question and 86 grams for the second. The second has to be solved by iteration (seeing what works)
Thanks a lot
A
Well, the amount of pyridoxine in those vitamins seems pretty uncertain, doesn't it? It's like the vitamins have their own personality, saying "I'm gonna have this amount of pyridoxine today, but maybe a different amount tomorrow!" Anyway, to figure out the probability of the vitamins containing less than 100 grams of pyridoxine, we can use some lovely statistics. First, let's calculate the z-score, which helps us measure how many standard deviations away from the mean the value is. In this case, the mean is 110 grams and the standard deviation is 25 grams. So, the z-score for 100 grams would be (100 - 110) / 25 = -0.4. Using a z-table or calculator, we can find that the probability of getting a value less than -0.4 is approximately 0.3446. Therefore, the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine is about 0.3446, or about 34.46%.
B
Ah, the elusive question of how much pyridoxine is needed in a vitamin for it to be considered a vitamin at all. Well, we know that approximately 83% of the vitamins will have at least that certain amount of pyridoxine. So, let's do some detective work. Since the distribution is normal, we can use a z-table or calculator to find the z-score that corresponds to 83% of the area under the curve. The z-score turns out to be about 0.9641. Now, to find out the corresponding value of pyridoxine, we can use the formula z = (x - μ) / σ, and rearrange it to solve for x. Plugging in the z-score of 0.9641, the mean of 110 grams, and the standard deviation of 25 grams, we find that x is approximately 134.1 grams. So, if you want at least 83% of the vitamins to have enough pyridoxine, you'll need a minimum of about 134.1 grams. But hey, don't worry, you can always add a little extra pyridoxine for good measure. After all, who doesn't love some extra vitamins?
A. To find the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine, we can use the standard normal distribution.
First, we need to calculate the z-score, which measures the number of standard deviations the value is from the mean. The formula for the z-score is:
z = (x - μ) / σ
where x is the value we're interested in, μ is the mean, and σ is the standard deviation. In this case, x = 100 grams, μ = 110 grams, and σ = 25 grams. Plug these values into the formula:
z = (100 - 110) / 25
z = -0.4
Next, we need to find the probability corresponding to this z-score. We can look up the z-score in a standard normal distribution table or use a statistical software/tool.
Using a standard normal distribution table, we find that the probability corresponding to a z-score of -0.4 is approximately 0.3446.
Therefore, the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine is approximately 0.3446.
B. To find the minimum amount of pyridoxine in grams that corresponds to approximately 83% of the vitamins, we need to find the z-score corresponding to the given probability.
We can use the standard normal distribution table or a statistical software/tool to find the z-score that corresponds to the given probability. In this case, the given probability is approximately 0.83.
Using a standard normal distribution table, we find that the z-score corresponding to a probability of 0.83 is approximately 0.952.
Now, we can calculate the value in grams using the z-score formula:
z = (x - μ) / σ
Since we want to find the minimum amount of pyridoxine, we can rearrange the formula to solve for x:
x = μ + (z * σ)
x = 110 + (0.952 * 25)
x = 133.8
Therefore, approximately 83% of the vitamins will have at least 133.8 grams of pyridoxine.