Question

1: Suppose a team of biologists has been studying the Pinedale children's fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists have determined that x has a normal distribution with mean u = 10.2 inches and standard deviation o = 1.4 inches. 1.1: What is the probability that a single trout taken at random from the pond is between 8.1 and 11.6 inches long? (5 Points) Solution 1.1: 1.2: Find the shortest and longest length of trout by the middle 80% of the data. (5 Points) Solution 1.2:

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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 ) = 10.2
standard deviation ( sd )= 1.4
----------------------------------------------------------------------------------------
(a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8.1) = (8.1-10.2)/1.4
= -2.1/1.4 = -1.5
= P ( Z <-1.5) From Standard Normal Table
= 0.0668
P(X < 11.6) = (11.6-10.2)/1.4
= 1.4/1.4 = 1
= P ( Z <1) From Standard Normal Table
= 0.8413
P(8.1 < X < 11.6) = 0.8413-0.0668 = 0.7745
----------------------------------------------------------------------------------------
(b)
LOWER/BELOW
P ( Z < x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is -0.841621
P( x-u/s.d < x - 10.2/1.4 ) = 0.2
That is, ( x - 10.2/1.4 ) = -0.841621
--> x = -0.841621 * 1.4 + 10.2 = 9.02173
UPPER/TOP
P ( Z > x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is 0.841621
P( x-u / (s.d) > x - 10.2/1.4) = 0.2
That is, ( x - 10.2/1.4) = 0.841621
--> x = 0.841621 * 1.4+10.2 = 11.37827
the shortest is 9.022
the longest is 11.38