A variable has a mean of 100 and a standard deviation of 6.  Four observations of this variable have a mean of 119 and a sample standard deviation of 8. z= , t =

To calculate the **z**-score and **t**-score, we need to clarify which of the two sets of data you're referencing.

### 1. **Z-Score**: 
The **z-score** measures how many standard deviations an observation is from the population mean. It is used when the population standard deviation is known.

The formula for the **z-score** is:

\[
z = \frac{X - \mu}{\sigma}
\]

Where:
- \( X \) is the sample mean
- \( \mu \) is the population mean
- \( \sigma \) is the population standard deviation

In your case:
- \( X = 119 \) (sample mean)
- \( \mu = 100 \) (population mean)
- \( \sigma = 6 \) (population standard deviation)

Substitute the values:

\[
z = \frac{119 - 100}{6} = \frac{19}{6} \approx 3.17
\]

So, the **z-score** is approximately **3.17**.

### 2. **T-Score**: 
The **t-score** is used when the population standard deviation is unknown, and we only have the sample standard deviation. It is computed using:

\[
t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}
\]

Where:
- \( X = 119 \) (sample mean)
- \( \mu = 100 \) (population mean)
- \( s = 8 \) (sample standard deviation)
- \( n = 4 \) (sample size)

Now substitute the values:

\[
t = \frac{119 - 100}{\frac{8}{\sqrt{4}}} = \frac{19}{\frac{8}{2}} = \frac{19}{4} = 4.75
\]

So, the **t-score** is **4.75**.

### Summary:
- **z-score**: \( \approx 3.17 \)
- **t-score**: \( 4.75 \)

Let me know if you'd like more details or have any further questions!

### 5 related questions:
1. What is the difference between a z-score and a t-score?
2. When should you use a t-distribution instead of a normal distribution?
3. How does sample size affect the t-distribution?
4. What happens to the t-score if the sample size increases?
5. Can we use a z-score when the population standard deviation is unknown?

### Tip:
For small sample sizes (typically \(n \leq 30\)), use the t-distribution since it accounts for more uncertainty in estimating the population parameters.

A variable has a mean of 100 and a standard deviation of 6. Four observations of this variable have a mean of 119 and a sample standard deviation of 8. z= , t =

Learn how to calculate the z-score and t-score for a variable with a population mean of 100 and a standard deviation of 6. Explore detailed step-by-step solutions for sample data with a mean of 119 and sample standard deviation of 8. This problem covers key statistical concepts, including z-distribution and t-distribution. Suitable for Grades 10-12.

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