hat percentage of the time will a variable that has a t distribution with the specified degrees of freedom fall in the indicated region? (Use a table or technology. Round your answers to one decimal place.) (a) 10 df, between −1.81 and 1.81 % (b) 21 df, between −2.08 and 2.08 % (c) 21 df, to the right of 2.08 %
a.
The random variable follows a t-distribution.
The degrees of freedom is, df = 10.
Thus, the probability that the random variable takes a value between –1.81 and 1.81 is:
P (–1.81 < tdf < 1.81)
= P (–1.81 < t10 < 1.81)
= P (t10 < 1.81) – P (t10 < 1.81).
Using Excel formula: =T.DIST(1.81,10,TRUE), P (t10 < 1.81) ≈ 0.9498.
Using Excel formula: =T.DIST(-1.81,10,TRUE), P (t10 < –1.81) ≈ 0.0502.
Thus,
P (–1.81 < t10 < 1.81)
= 0.9498 – 0.0502
= 0.8996.
The percentage is the probability multiplied by 100, which in this case is, 90.0% (= 0.8996 × 100, correct to one decimal place).
Thus, the required percentage is 90.0%.
b.
The degrees of freedom is, df = 21.
Thus, the probability that the random variable takes a value between –2.08 and 2.08 is:
P (–2.08 < tdf < 2.08)
= P (–2.08 < t21 < 2.08)
= P (t21 < 2.08) – P (t21 < 2.08).
Using Excel formula: =T.DIST(2.08,21,TRUE), P (t21 < 2.08) ≈ 0.9750.
Using Excel formula: =T.DIST(-2.08,21,TRUE), P (t21 < –2.08) ≈ 0.0250.
Thus,
P (–2.08 < t21 < 2.08)
= 0.9750 – 0.0250
= 0.9500.
The percentage is the probability multiplied by 100, which in this case is, 95.0% (= 0.9500 × 100).
Thus, the required percentage is 95.0%.
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