Mathematics, 07.04.2020 01:05, Jasten

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. Let X be the number of days for a randomly selected trial. Round all answers to two decimal places. A. X ~ N( , )
B. If one of the trials is randomly chosen, find the probability that it lasted at least 21 days.
C. If one of the trials is randomly chosen, find the probability that it lasted between 15 and 20 days.
D. 85% of all of these types of trials are completed within how many days?

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answered: drhtgsehg6334

a) $X \sim N(18,5)$

b) $P(X21)=P(\frac{X-\mu}{\sigma}\frac{21-\mu}{\sigma})=P(Z0.6)$

And we can find this probability with the complement rule and with the normal standard table or excel and we got:

P(z0.6)=1-P(z

c) P(15

And we can find this probability with this difference and with the normal standard table or excel and we got:

P(-0.6

d) For this case we can use the definition of z score, since we need 85% of the values in the middle of the distribution on the tails we need the remaining 1-0.85 = 0.15 and the value for $\alpha/2 = 0.15/2= 0.075$ and if we look that accumulates 0.075 of the area on the left of the normal standard distribution we got:

z = -1.440

And using the z score formula we got:

$-1.440 = \frac{X -18}{5}$

And solving for X we got : X= 18 -1.440*5 = 10.8

And since the distribution is symmetrical we got:

$1.440 = \frac{X -18}{5}$

And solving for X we got : X= 18 +1.440*5 = 25.2

then we have 85% of the values between (10.8; 25.2)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the duration of a particular type of criminal of a population, and for this case we know the distribution for X is given by:

$X \sim N(18,5)$

Where $\mu=18$ and $\sigma=5$

Part b

We are interested on this probability

P(X21)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X21)=P(\frac{X-\mu}{\sigma}\frac{21-\mu}{\sigma})=P(Z0.6)$

And we can find this probability with the complement rule and with the normal standard table or excel and we got:

P(z0.6)=1-P(z

Part c

P(15

And we can find this probability with this difference and with the normal standard table or excel and we got:

P(-0.6

Part d

For this case we can use the definition of z score, since we need 85% of the values in the middle of the distribution on the tails we need the remaining 1-0.85 = 0.15 and the value for $\alpha/2 = 0.15/2= 0.075$ and if we look that accumulates 0.075 of the area on the left of the normal standard distribution we got: