Question

suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 18 days and a standard deviation of 5 days. a. if one of the trials is randomly chosen, find the probability that it lasted at least 21 days. select b. if one of the trials is randomly chosen, find the probability that it lasted between 15 and 20 days. select c. what is the z - score of a trial that lasts 13 days? select

Explanation:

Step1: Define given parameters

Mean $\mu = 18$, standard deviation $\sigma = 5$

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Part A

Step2: Calculate z-score for 21 days

$z = \frac{x - \mu}{\sigma} = \frac{21 - 18}{5} = 0.6$

Step3: Find $P(X \geq 21)$

This equals $1 - P(Z < 0.6)$. Using z-table, $P(Z < 0.6) = 0.7257$, so:
$1 - 0.7257 = 0.2743$

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Part B

Step4: Calculate z-scores for 15 and 20

For $x=15$: $z_1 = \frac{15 - 18}{5} = -0.6$
For $x=20$: $z_2 = \frac{20 - 18}{5} = 0.4$

Step5: Find $P(15 < X < 20)$

This equals $P(Z < 0.4) - P(Z < -0.6)$. Using z-table: $P(Z < 0.4)=0.6554$, $P(Z < -0.6)=0.2743$, so:
$0.6554 - 0.2743 = 0.3811$

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Part C

Step6: Calculate z-score for 13 days

$z = \frac{x - \mu}{\sigma} = \frac{13 - 18}{5} = -1$

Answer:

A. 0.2743
B. 0.3811
C. -1