Answer:
Step-by-step explanation:
Given that the uniform distribution on the interval (8.5, 19) as a model for depth (cm) of the bioturbation layer in sediment in a certain region
a) Mean =[tex]\frac{1}{19-8.5} =13.75[/tex]
b) Variance = [tex]\frac{19^2-8.5^2}{12} \\=24.0625[/tex]=24.06
c) the probability that observed depth is at most 10=[tex]\frac{10-8.5}{19-8.5} =0.1429[/tex]
d) the probability that observed depth is between 10 and 15=
[tex]\frac{15-10}{19-8.5} =0.4762[/tex]
e) the probability that the observed depth is within 1 standard deviation of the mean value
=P(13.75-4.9053<x<13.75+4.9053)
=P(8.8447<x<18.7553)
= P(8.84<x<18.76)
[tex]\frac{9.92}{10.5} \\=0.945[/tex]
f) 2 std deviation will cover full interval
So prob =1
