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Use the ALEKS calculator to solve the following problems.

(a) Consider a [tex]t[/tex] distribution with 14 degrees of freedom. Compute [tex]P(-1.31 \ \textless \ t \ \textless \ 1.31)[/tex]. Round your answer to at least three decimal places.

[tex]\[ P(-1.31 \ \textless \ t \ \textless \ 1.31) = \square \][/tex]

(b) Consider a [tex]t[/tex] distribution with 2 degrees of freedom. Find the value of [tex]c[/tex] such that [tex]P(t \geq c) = 0.01[/tex]. Round your answer to at least three decimal places.

[tex]\[ c = \square \][/tex]


Sagot :

Sure, let's solve these problems step-by-step.

### Part (a)

Consider a [tex]\( t \)[/tex]-distribution with 14 degrees of freedom. We are asked to compute [tex]\( P(-1.31 < t < 1.31) \)[/tex].

1. We need to find the cumulative probability of [tex]\( t \)[/tex]-values between -1.31 and 1.31 for a [tex]\( t \)[/tex]-distribution with 14 degrees of freedom.
2. This involves finding the cumulative distribution function (CDF) values at [tex]\( t = 1.31 \)[/tex] and [tex]\( t = -1.31 \)[/tex], then finding the difference between these CDF values.

The CDF value at [tex]\( t = 1.31 \)[/tex] for 14 degrees of freedom is approximately 0.894.
The CDF value at [tex]\( t = -1.31 \)[/tex] for 14 degrees of freedom is approximately 0.105.

Now, calculate [tex]\( P(-1.31 < t < 1.31) \)[/tex] by subtracting:
[tex]\[ P(-1.31 < t < 1.31) = \text{CDF}(1.31) - \text{CDF}(-1.31) \][/tex]
[tex]\[ P(-1.31 < t < 1.31) = 0.894 - 0.105 \][/tex]
[tex]\[ P(-1.31 < t < 1.31) = 0.789 \][/tex]

So, the answer is:
[tex]\[ P(-1.31 < t < 1.31) = 0.789 \][/tex]

### Part (b)

Consider a [tex]\( t \)[/tex]-distribution with 2 degrees of freedom, and we need to find the value of [tex]\( c \)[/tex] such that [tex]\( P(t \geq c) = 0.01 \)[/tex].

1. To solve this, find the critical value [tex]\( c \)[/tex] where the upper tail probability equals 0.01 for 2 degrees of freedom.
2. This means we need to find the inverse cumulative distribution function value (also known as the percent point function, PPF) for the upper 1% tail.

The value of [tex]\( c \)[/tex] where [tex]\( P(t \geq c) = 0.01 \)[/tex] is approximately 6.965.

So, the answer is:
[tex]\[ c = 6.965 \][/tex]

Summary:
(a) [tex]\( P(-1.31 < t < 1.31) = 0.789 \)[/tex]
(b) [tex]\( c = 6.965 \)[/tex]