Question

For a t distribution with 18 degrees of freedom, find the area, or probability, in each region. (Round your answers to three decimal places.) (a) to the right of 1.734 (b) to the left of 2.101 (C) to the left of -2.552 (d) to the right of 1.330 (e) between – 1.734 and 1.734 (f) between – 2.552 and .552

Answer 1

Solution degrees of freedom = 18 (a) P(t > 1.734) = 0.050 Therefore the to the right Probability of 1.734 is 0.050

(6) P(t <2.100) =HP p(t >2.101) = -0.025 0.975 to the left of 2.106 is (0.975 Therefore the Porobability P(t <-2.552) Therefore the Probability 0.010 left of -2.552 is o.olo (d) PA> 1.330) = 0.100 1.330 is 0.100 Therefore the probability to the right of (e) p(-6756 I-0.100 1.734 <t< 1.734) 0.900 between and 1.734 - 1.734 Therefore the is 0.900 probability (A) P (-23 2.552 <t<2.552) 1-0.020 = 0.980 between -2.552 and 2.552 is Therefore the probability 0.980