SOLUTION: The annual sales of dealers of a company follows the normal distribution with its mean as K94. A random sample of 10 dealers of the company is taken from the normal population. T

Question 1183988: The annual sales of dealers of a company follows the normal distribution with its mean as
K94. A random sample of 10 dealers of the company is taken from the normal population.
The variance of the annual sales of these 10 dealers is K81. Find the probability that the
mean annual sales of the sample is
i. Less than K98
ii. More than K98
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**1. Identify the given information:**
* Population mean (μ) = K94
* Sample size (n) = 10
* Sample variance (s²) = K81
* Sample standard deviation (s) = √81 = K9
**2. Calculate the standard error of the mean:**
The standard error of the mean (SEM) measures the variability of sample means. Because the population standard deviation is unknown, we use the sample standard deviation to estimate it and use the t-distribution.
SEM = s / √n = 9 / √10 ≈ 2.846
**3. Calculate the t-score:**
The t-score tells us how many standard errors the sample mean is away from the population mean.
t = (sample mean - population mean) / SEM
* **i. Less than K98:**
t = (98 - 94) / 2.846 ≈ 1.406
* **ii. More than K98:**
The t-score will be the same magnitude but positive: 1.406
**4. Find the probabilities using the t-distribution:**
Since the sample size is small (n < 30), we use the t-distribution with (n-1) = 9 degrees of freedom. You'll need a t-table or a statistical calculator for this step.
* **i. Less than K98:** Look up the probability associated with a t-score of 1.406 and 9 degrees of freedom. This gives you the area to the *left* of 98. A t-table or calculator will show it is approximately 0.90. This means there is a 90% chance of the sample mean being less than K98.
* **ii. More than K98:** Since the t-distribution is symmetric, the probability of the sample mean being *more* than K98 is 1 minus the probability of it being less than K98. Therefore, the probability is 1 - 0.90 = 0.10. This means there is a 10% chance of the sample mean being greater than K98.
**Answer:**
* **i. Probability of sample mean less than K98:** Approximately 0.90 or 90%
* **ii. Probability of sample mean more than K98:** Approximately 0.10 or 10%

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