pent pent pent pent pent pent Name: MA 341-Fall 2016 Dr. Selby Homework 6- Due 1

Question

pent pent pent pent pent pent Name: MA 341-Fall 2016 Dr. Selby Homework 6- Due 11/21/2017 1. The average number of ladybugs in a group (called a "loveliness") is approximately normally distributed with a mean of 100,000 and a standard deviation of 10,000. a. Illustrate the distribution. Include a mean and two standard deviations from the mean. b. What is the probability of finding a loveliness with more than 130,000 ladybugs? c. What is the 45 percentile of ladybug loveliness sizes? d. What is the probability of finding a loveliness with between 88,971 and 111,111 lady bugs? 2. For each of the following, define the random variable, describe which distribution is likely the best fit, and n700fr1684948c5a789d4e793a9b266bb4e18d7a67983fe498732

Explanation / Answer

1.
a.
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 ) = 100000
standard Deviation ( sd )= 10000
two standard deviation = 2*10000 = 20000

b.
probability of finding a loveliness with more than 130,000 ladybugs
P(X > 130000) = (130000-100000)/20000
= 30000/20000 = 1.5
= P ( Z >1.5) From Standard Normal Table
= 0.0668

c. 45th percentile of ladybugs loveliness sizes
P ( Z < x ) = 0.45
Value of z to the cumulative probability of 0.45 from normal table is -0.1257
P( x-u/s.d < x - 100000/20000 ) = 0.45
That is, ( x - 100000/20000 ) = -0.1257
--> x = -0.1257 * 20000 + 100000 = 97486.7731

d. probability that finding loveliness with between 88971 and 111111
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 88971) = (88971-100000)/20000
= -11029/20000 = -0.5515
= P ( Z <-0.5515) From Standard Normal Table
= 0.2907
P(X < 111111) = (111111-100000)/20000
= 11111/20000 = 0.5556
= P ( Z <0.5556) From Standard Normal Table
= 0.7107
P(88971 < X < 111111) = 0.7107-0.2907 = 0.4201

2.
Random variable
The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values
normal distribution is best fit method solve and each problem sample mean X is a variable

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.