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1 answer:
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Hope this helps! This is how you do it.
Answer with Step-by-step explanation:
We are given that a sample space
S={a,b,c,d,e}
P(a)=0.1
P(b)=0.1
P(c)=0.2
P(d)=0.4
P(e)=0.2
a.A={a,b,c}
P(A)=P(a)+P(b)+P(c)
P(A)=0.1+0.1+0.2=0.4
b.B={c,d,e}
P(B)=P(c)+P(d)+P(e)=0.2+0.4+0.2=0.8
c.A'=Sample space-A={a,b,c,d,e}-{a,b,c}={d,e}
P(A')=P(d)+P(e)=0.4+0.2=0.6
d.={a,b,c,d,e}
=P(a)+P(b)+P(c)+P(d)+P(e)=0.1+0.1+0.2+0.4+0.2=1
e.={c}
Answer:
The probability of finding a sample mean less than 18 hours is 0.0082
Step-by-step explanation:
To find the probability of finding a sample mean less than 18 hours, we need to calculate the z-score of this sample mean 18. And the probability of finding a sample mean less than 18 hours is P(z<z(18)).
Z-score can be calculated as follows:
z(18)= where
- X is the sample mean (18 hours)
- M is the average hours dentists spend per week on fillings (20 hours)
- s is the standard deviation (10 hours)
- N is the sample size (144)
Putting the numbers, we get:
z(18)=
Using z- table we can find that P(z<z(18)) = 0.0082