Question

suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of 5 days. let x be the number of days for a randomly selected trial. round all answers to 4 decimal places where possible.
a. what is the distribution of x? x ~ n( , )
b. if one of the trials is randomly chosen, find the probability that it lasted at least 18 days.
c. if one of the trials is randomly chosen, find the probability that it lasted between 22 and 26 days.
d. 86% of all of these types of trials are completed within how many days? (please enter a whole number)
hint:
helpful videos:

• find a probability +
• finding a value given a probability +

hint

Explanation:

Step1: Identify normal - distribution parameters

Given mean $\mu = 21$ and standard deviation $\sigma=5$. The distribution of $X$ is $X\sim N(21,5)$.

Step2: Standardize for part b

We want $P(X\geq18)$. First, calculate the z - score $z=\frac{x-\mu}{\sigma}=\frac{18 - 21}{5}=\frac{- 3}{5}=-0.6$. Then $P(X\geq18)=1 - P(X < 18)=1 - P(Z < - 0.6)$. Using the standard normal table, $P(Z < - 0.6)=0.2743$, so $P(X\geq18)=1 - 0.2743 = 0.7257$.

Step3: Standardize for part c

For $P(22

Step4: Find the value for part d

We want to find $x$ such that $P(X < x)=0.86$. Looking up the z - score in the standard normal table corresponding to a probability of $0.86$, the z - score $z\approx1.08$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $1.08=\frac{x - 21}{5}$. Solving for $x$, we get $x=21+1.08\times5=21 + 5.4=26.4\approx26$.

Answer:

a. $21,5$
b. $0.7257$
c. $0.2620$
d. $26$