46%. Tue 8:04 PM Danielle D w) Statistical Analysis in Crimin 2- sign Layout Ref

Question

46%. Tue 8:04 PM Danielle D w) Statistical Analysis in Crimin 2- sign Layout References Mailings Review View 9Statistical Analysis in Criminal Justice Homework #6-Due 10/18 4:00pm Nam Please answer the following on this page or on additional paper if needed. You must show ALL of your work to receive full credit for your answers. Your final answer for math problems must be enclosed in a box. Your homework MUST BE COMPLETED INDEPENDENTLY OR IT IS CONSIDERED ACADEMIC DISHONESTY 1. According to the 2005 Census of Publicly Funded Forensic Crime Laboratories, state crime labs received a mean of 610 requests for DNA analyses during the year, with a population standard deviation of 581. Illinois State Police Crime Lab received 475 DNA requests during 2005? What was illinois State Police Crime Lab's z-score? a. b. What proportion of states had more than 650 requests during 2005? c. What percentage of states had between 450 and 600 requests during 2005? d. What is the minimum number of requests a state crime lab has to receive to be in the top 10% in requests?

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 610
standard Deviation ( sd )= 581
I.
P(X = 475) = (475-610)/581
= -135/581= -0.2324
II.
P(X > 650) = (650-610)/581
= 40/581 = 0.0688
= P ( Z >0.0688) From Standard Normal Table
= 0.4726
the proportion f receive more than 650 is 47.26%
III.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 450) = (450-610)/581
= -160/581 = -0.2754
= P ( Z <-0.2754) From Standard Normal Table
= 0.3915
P(X < 600) = (600-610)/581
= -10/581 = -0.0172
= P ( Z <-0.0172) From Standard Normal Table
= 0.4931
P(450 < X < 600) = 0.4931-0.3915 = 0.1016
the proportion of 450 to 600 = 10.16%
IV.
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.2816
P( x-u / (s.d) > x - 610/581) = 0.1
That is, ( x - 610/581) = 1.2816
--> x = 1.2816 * 581+610 = 1354.5815 ~ 1355
the no.of request to receive in top should be atleast 1355

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