Question

1)Today, the waves are crashing onto the beach every 5.6 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.6 seconds. Round to 4 decimal places where possible.

a. The mean of this distribution is

b. The standard deviation is  

c. The probability that wave will crash onto the beach exactly 0.4 seconds after the person arrives is P(x = 0.4) =  

d. The probability that the wave will crash onto the beach between 1.8 and 3.6 seconds after the person arrives is P(1.8 < x < 3.6) =  

e. The probability that it will take longer than 2.02 seconds for the wave to crash onto the beach after the person arrives is P(x > 2.02) =  

f. Suppose that the person has already been standing at the shoreline for 0.2 seconds without a wave crashing in. Find the probability that it will take between 1.6 and 2.5 seconds for the wave to crash onto the shoreline.

g. 74% of the time a person will wait at least how long before the wave crashes in? seconds.

h. Find the minimum for the upper quartile.  seconds.

2)Suppose that the weight of an newborn fawn is Uniformly distributed between 1.5 and 3.2 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible.

a. The mean of this distribution is

b. The standard deviation is

c. The probability that fawn will weigh exactly 1.8 kg is P(x = 1.8) =  

d. The probability that a newborn fawn will be weigh between 2.2 and 2.7 is P(2.2 < x < 2.7) =  

e. The probability that a newborn fawn will be weigh more than 2.64 is P(x > 2.64) =

f. P(x > 1.7 | x < 1.9) =

g. Find the 34th percentile.

Answer 1

This is a uniform distribution with

Since we know that
Probability density function of a uniform distribution is

This implies that
Cummulative density function of a uniform distribution is

a)Since we also know that
Mean of a uniform distribution is the average of its interval i.e.

Mean = 2.8
b) Also

Variance = 2.6133

Standard Deviation = 1.6166
c) For a continous distribution probability at a particular point is equal to 0

Therefore, P(x=0.4) = 0
d) Pr(x1<X<x2) = F(x2) - F(x1)
Where x1 = 1.8
x2 = 3.6

e) Pr(X>x) = 1- F(x)
Where x = 2.02

Pr(X>2.02) = 0.6393