asbestos standards
stats please show work to explain how?
Suppose the length of time to onset of lung cancer in asbestos workers is approximately normally distributed with a mean of 26 years and a variance of 4 years squared.
a. What is the standard deviation?
b. What interval of onset times includes about 95% of the workers?
Approximate using empirical rule.
d. What is the probability that an asbestos worker gets lung cancer sooner than 22 years?
please show work so that i can use the method to figure out similar questions thank you
I think you mean 16 when you say 4 squared. Variance is 16.
a) Standard deviation = sqrt(variance) = sqrt(16)=4
b) From the normal probability table 95 % of cases are included between -1.96 and 1.96. We need to convert these standardized values to actual values.
z= (x-mean) / SD
-1.96 = (x-26)/4
solve for x
x=(4)(-1.96)+26
x= -7.84+26 = 18.16
1.96 = (x-26) / 4
x=(4)(1.96)+26=33.84
The interval (18.16, 33.84) of onset times includes about 95% of the workers.
d) P( x < 22)
μ = 26
σ = 4
P(x < 22) = P( z < (22-26) / 4)
= P(z < -1) = 0.1587
(from normal probability table)
Ikarbus – IK103/IK103 CNG
