Jumbo eggs in Australia, on average, are supposed to weigh 68g. Margot is in charge of a quality control test that involves weig
2 answers:
Answer:
0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g
Step-by-step explanation:
To solve this question, we use the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean weight of 67g and a sample standard deviation of 1g.
This means that
Sample of 4
This means that
What is the probability that the mean weight of 4 eggs in a package is less than 68.5g?
This is the pvalue of Z when X = 68.5. So
By the Central Limit Theorem
has a pvalue of 0.9987
0.9987 = 99.87% probability that the mean weight of 4 eggs in a package is less than 68.5g
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Answer:
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4
Step-by-step explanation:
The given function is presented as follows;
From the given function, we have;
When x = 1, the denominator of the fraction, , which is (x - 1) = 0, and the function becomes,
therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote
Also we have that in the given function, as <em>x</em> increases, the fraction tends to 0, therefore as x increases, we have;
Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1
When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, . When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left
Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.