(b) Consider a t distribution with 22 degrees of freedom. Find the value- of $c$ such that $P(1 \leq c)=0.10$. Round your-answer to at least three decimal places.
\[
c=\mathbb{\pi}
\]
Answer
Final Answer: The value of \(c\) such that \(P(T \leq c)=0.10\) for a t-distribution with 22 degrees of freedom is approximately \(\boxed{-1.321}\).
Step 1 ：The question is asking for the value of c such that the probability of the t-distribution with 22 degrees of freedom being less than or equal to c is 0.10. This is equivalent to finding the 10th percentile of the t-distribution with 22 degrees of freedom.
Step 2 ：We can use the Percent Point Function (PPF), which is the inverse of the Cumulative Distribution Function (CDF), to find this value. The PPF gives the value of the random variable such that the probability of the variable being less than or equal to that value equals the given probability.
Step 3 ：Let's denote the degrees of freedom as \(df = 22\) and the probability as \(p = 0.1\).
Step 4 ：By using the PPF, we find that \(c = -1.3212367416133615\).
Step 5 ：This means that the probability of the t-distribution with 22 degrees of freedom being less than or equal to -1.321 is 0.10.
Step 6 ：Final Answer: The value of \(c\) such that \(P(T \leq c)=0.10\) for a t-distribution with 22 degrees of freedom is approximately \(\boxed{-1.321}\).
