Complete question:
A manufacturer has determined that a model of its washing machine has an expected life that is Exponential with a mean of four years to failure and irrelevant board burn-in period. He wants to testthe system and complete data collection. Find the probability that one of these washing machines will have a life that ends: (Note you can find the reliability of the washing machine life)
a) After an initial four years of washing machine service
b) Before four years of washing machine service are completed
c) Not before six years of washing machine service.
Answer:
a) 0.3679
b) 0.6321
c) 0.2231
Step-by-step explanation:
Given:
Mean, u= 4
/\ = 1/u
= 1/4 = 0.25
The cummulative distribution function, will be:
For x≥0,


a) After an intial four years:

P(x>4) = 0.3679
b) Before four years:

P(x<4) = 0.6321
c) Not before 6 years:

P(x>6) = 0.2231
Answer:
n-185
Step-by-step explanation:
since 185 needs to be subtracted from a number, or n
Step-by-step explanation:
1)......mean= sum of observations/ total observations
= 5.2+8.4+4.3+6.7+5.8/5= 6.08
median= n=5(odd)=n+1/2=6/2=3rdterm= 5.8( when arranged in ascending order).....
mode= 3median- 2meam= 3×5.8-2×6.08= 5.24
range=8.4-4.3=4.1(when done A.O)
2)......mean= -2-13-13-5-7/5= -8
median= 3rd term= -5.....(n=5..odd)(when done A.o)
mode=largest frequency= -13
range= -13-(-2) = -13+2= -11(A.O)
3).....mean= sum/no.= 8+6+13/2+17/2+15/2+7+8+11/2+7+8//10=7.2
median=n=10(even)
=
mode= 8
range=8.5-5.5=3
median:==
(10/2)+(10/2+1)//2=5 term + 6term/2=7+7.5=14.5/2=7.25
4)....
5)...remove 62...then 73will be having highest frequency
Answer:
$7.57
Step-by-step explanation:
$0.04 x 20 = 0.80
0.80 + 6.77 = 7.57