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the distribution for the length of a person's tongue is normally distributed. a random sample of 3 people had an average of 6.9 inches with a standard deviation of 1.1 inches. find a 92% confidence interval for the true average length of a tongue.

User Mihir Dave
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Given a sample size of 3 people with a sample mean tongue length of 6.9 inches and a sample standard deviation of 1.1 inches, we want to estimate the population mean. We'll do this by calculating a 92% confidence interval.

Step 1: Find the z critical value (also known as z* value).
This is the number of standard deviations from the mean we'd need to move to capture the middle 92% of the data in a normal distribution. After looking it up in a z-table or calculator, we find that the z critical value for a 92% confidence level is approximately 1.751.

Step 2: Compute the margin of error.
The margin of error is calculated using the following formula:

Margin of Error = z* * (standard deviation / sqrt(sample size))

Here, the standard deviation is 1.1, sample size is 3 and our z* value is 1.751.

So, Margin of Error = 1.751 * (1.1 / sqrt(3)) = 1.112 (rounded to three decimal places).

Step 3: Determine the confidence interval.
Now we come to calculating the confidence interval. We do this by subtracting the margin of error from the sample mean and also adding it to the sample mean:

Lower bound = sample mean - margin of error = 6.9 - 1.112 = 5.788

Upper bound = sample mean + margin of error = 6.9 + 1.112 = 8.012

Hence, based on our sample data, we are 92% confident that the length of an average person's tongue lies between 5.788 inches and 8.012 inches.

User Flappix
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