Part (a)

You need to calculate the probability $P(t \leq -1.51)$ for a t-distribution with 24 degrees of freedom.

For a t-distribution with 24 degrees of freedom, you can use the cumulative distribution function (CDF) to find this probability.

$P(t \leq -1.51) \approx 0.071$

Part (b)

You need to find the value of $c$ such that $P(-c < t < c) = 0.90$ for a t-distribution with 5 degrees of freedom.

This means you are looking for the t-value where the middle 90% of the distribution lies.

For a t-distribution with 5 degrees of freedom, the critical value $c$ can be found using the t-table or a statistical calculator. The value of $c$ is:

$c \approx 2.015$

Summary:

• (a) $P(t \leq -1.51) \approx 0.071$
• (b) $c \approx 2.015$

Would you like more details or have any questions about these calculations? Here are some related questions:

1. How do degrees of freedom affect the shape of a t-distribution?
2. What is the difference between a t-distribution and a normal distribution?
3. How do you interpret the result from part (a) in a practical context?
4. Can you explain the concept of the confidence interval related to part (b)?
5. How would the value of $c$ change if the confidence level was 95% instead of 90%?

Tip: The t-distribution is particularly useful when dealing with small sample sizes, as it accounts for the extra variability introduced by estimating the population standard deviation.