The random sample shown below was selected from a normal distribution. 8, 10, 7, 5, 3, 3 D Complete parts a and b.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![### Confidence Interval Calculation for Population Mean
**Problem Statement:**
The random sample shown below was selected from a normal distribution:
8, 10, 7, 5, 3, 3
Complete parts a and b.
**a. Construct a 99% confidence interval for the population mean \( \mu \).**
**Solution:**
**Step 1: Calculate the Sample Mean (\( \overline{x} \))**
First, we sum the sample data and then divide by the number of data points:
\[ \overline{x} = \frac{8 + 10 + 7 + 5 + 3 + 3}{6} = \frac{36}{6} = 6 \]
**Step 2: Calculate the Sample Standard Deviation (s)**
Next, compute the variance by finding the squared differences from the mean, summing them, and then dividing by the number of observations minus one.
\[ s^2 = \frac{(8-6)^2 + (10-6)^2 + (7-6)^2 + (5-6)^2 + (3-6)^2 + (3-6)^2}{6 - 1} \]
\[ s^2 = \frac{(2)^2 + (4)^2 + (1)^2 + (-1)^2 + (-3)^2 + (-3)^2}{5} \]
\[ s^2 = \frac{4 + 16 + 1 + 1 + 9 + 9}{5} = \frac{40}{5} = 8 \]
\[ s = \sqrt{8} \approx 2.83 \]
**Step 3: Find the Critical Value**
For a 99% confidence interval and degrees of freedom \( n-1 = 6-1 = 5 \), we use the t-distribution. The critical value from the t-table for 5 degrees of freedom at 99% confidence is approximately 4.032.
**Step 4: Calculate the Margin of Error (E)**
\[ E = t_{critical} \times \frac{s}{\sqrt{n}} \]
\[ E = 4.032 \times \frac{2.83}{\sqrt{6}} = 4.032 \times 1.156 \approx 4.66 \]
**Step 5: Construct the Confidence Interval**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb25aa5e3-8edb-4708-bfd5-abf65dd8da34%2Fd97e5cc5-ef1c-444b-b1a5-5096440d533c%2Fxkcogxa_processed.jpeg&w=3840&q=75)
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