Let's solve the given problems step-by-step:
(a) Compute [tex]\(P(t \leq -1.2)\)[/tex] for a [tex]\(t\)[/tex]-distribution with 5 degrees of freedom:
1. We need to find the cumulative probability for a [tex]\(t\)[/tex]-distribution with 5 degrees of freedom at [tex]\(t = -1.2\)[/tex].
2. Using the t-distribution table or a statistical tool, we find that [tex]\(P(t \leq -1.2)\)[/tex] for 5 degrees of freedom is approximately 0.142 (rounded to three decimal places).
Therefore,
[tex]\[ P(t \leq -1.2) = 0.142 \][/tex]
(b) Find the value of [tex]\(c\)[/tex] for a [tex]\(t\)[/tex]-distribution with 15 degrees of freedom such that [tex]\(P(-c < t < c) = 0.95\)[/tex]:
1. We are given that 95% of the data falls between [tex]\(-c\)[/tex] and [tex]\(c\)[/tex]. This implies the middle 95% of the [tex]\(t\)[/tex]-distribution, leaving 2.5% in each tail (since [tex]\(100\% - 95\% = 5\%\)[/tex], and [tex]\(5\%/2 = 2.5\%\)[/tex]).
2. From the t-distribution table or using a statistical tool, we find the t-value that corresponds to the cumulative probability of 0.975 (since we want the value of [tex]\(c\)[/tex] such that the central 95% of the distribution lies between [tex]\(-c\)[/tex] and [tex]\(c\)[/tex]). For 15 degrees of freedom, this is approximately 2.131 (rounded to three decimal places).
Therefore,
[tex]\[ c = 2.131 \][/tex]
In conclusion:
(a) [tex]\( P(t \leq -1.2) = 0.142 \)[/tex]
(b) [tex]\( c = 2.131 \)[/tex]
