Example Question - mean and standard deviation
Here are examples of questions we've helped users solve.
Calculating Specific Expansion Terms and Investigating Geometric Mean Relation
For the 2nd question (problem 8), since it is cut off, we are unable to provide a complete solution. However, for the 3rd question (problem 9), we can proceed: <p>The sum of the two numbers is $6$ times their geometric mean, then </p> <p>Let the two numbers be $a$ and $b$. Given that $a+b = 6\sqrt{ab}$ and we are to show that $a$ and $b$ are in the ratio $(3 + 2\sqrt{2}):(3 - 2\sqrt{2})$.</p> <p>Since we have a ratio of sums to a geometric mean, let's assume $a = (3 + 2\sqrt{2})k$ and $b = (3 - 2\sqrt{2})k$, where $k$ is some positive constant.</p> <p>\[ a + b = (3 + 2\sqrt{2})k + (3 - 2\sqrt{2})k\]</p> <p>\[ a + b = 3k + 2\sqrt{2}k + 3k - 2\sqrt{2}k\]</p> <p>\[ a + b = 6k\]</p> <p>The geometric mean of $a$ and $b$ is $\sqrt{ab}$:</p> <p>\[\sqrt{ab} = \sqrt{(3 + 2\sqrt{2})k \cdot (3 - 2\sqrt{2})k}\]</p> <p>\[\sqrt{ab} = \sqrt{(9 - 8)k^2}\]</p> <p>\[\sqrt{ab} = k\]</p> <p>Now, we compare the sum and geometric mean:</p> <p>\[6k = 6\sqrt{ab}\]</p> <p>\[k = \sqrt{ab}\]</p> <p>This confirms that the sum of $a$ and $b$ is indeed $6$ times their geometric mean, and thus the numbers are in the desired ratio.</p> For question 9, which discusses mean and standard deviation, no specific mathematical equations are shown, and thus a solution cannot be provided based on the information given in the image.
Understanding Normal Distribution Probabilities
This image presents a series of questions related to a normal distribution with a mean (μ) of 50 and a standard deviation (σ) of 4. The problem refers to a sample (n) of 100. The tasks are to find probabilities for certain values of X, which seems to represent a random variable. Here's how each question is addressed: a. The probability that X is less than 49: From the image, it looks like the probability, `P(X < 49)`, is given as 0.0062. b. The probability that X is between 49 and 50.5: The probability, `P(49 < X < 50.5)`, appears to be 0.8882. c. The probability that X is above 50.6: The probability, `P(X > 50.6)`, is shown as 0.0688. d. The value above which there is a 40% chance that X falls: The value of X, such that `P(X > X_value) = 0.40`, is given as 50.1012. These probabilities might have been calculated using a standard normal distribution table (Z-table) or a software that computes these values based on the Z-score formula. The Z-score is calculated by taking the difference between a value and the mean, then dividing by the standard deviation. In the case of part d, this Z-score would correspond with the 60th percentile value when looking up the normal distribution (since having 40% above the value means 60% is below it), and then this Z-score would be converted back to the specific value using the mean and standard deviation of the distribution.
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