Answer:
0.0000
Unusual
Step-by-step explanation:
Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.
i.e. population parameters are
The approximate probability that the sample meanwould have been as high or higher than 3.1
=0.0000
Answer:
0%
Step-by-step explanation:
We have the following information:
mean (m) = 208 days
standard deviation (s) = 15 days
We have that the z value would be:
z = (x - m) / sd
replacing we have:
z = (308 - 208) / 15
z = 6.66
However,
P (x> 308) = 1 - (x <308)
P (x> 308) = 1 - (z <6.66)
if we look for this value of z in the table we assume that it is 1, since the table reaches 3 and the value is 1. Therefore:
P (x> 308) = 1 - 1
P (x> 308) = 0
Which means that the probability is 0%, therefore it is a very unusual pregnancy.
