The life expectancy in Canada is 75 years with a standard deviation of 7 years.
ID: 3365548 • Letter: T
Question
The life expectancy in Canada is 75 years with a standard deviation of 7 years. A random sample of 100 individuals is taken.
In blank #1 enter the mean of the sample means.
In blank #2 enter the standard deviation of the sample means. (Round your answer to 1 decimal places and enter your answer as 0.1 not .1.)
Is the distribution of the sample means normal? Enter "yes" or "no" in blank #3.
In blank #4 enter the probability that the sample mean is less than 73.7 years. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
In blank #5 enter the probability that the sample mean is greater than 76.5 years. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
In blank #6 enter the probability that the sample mean is between 74.5 and 76 years. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
mean of the sampling distribution ( x ) = 75
b.
standard Deviation ( sd )= 7/ Sqrt ( 100 ) =0.7
sample size (n) = 100
c.
sample of 100 individuals is taken, it is large enough to conclude sample is normal
d.
P(X < 73.7) = (73.7-75)/7/ Sqrt ( 100 )
= -1.3/0.7= -1.8571
= P ( Z <-1.8571) From Standard NOrmal Table
= 0.0316
e.
P(X > 76.5) = (76.5-75)/7/ Sqrt ( 100 )
= 1.5/0.7= 2.1429
= P ( Z >2.1429) From Standard Normal Table
= 0.0161
f.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 74.5) = (74.5-75)/7/ Sqrt ( 100 )
= -0.5/0.7
= -0.7143
= P ( Z <-0.7143) From Standard Normal Table
= 0.2375
P(X < 76) = (76-75)/7/ Sqrt ( 100 )
= 1/0.7 = 1.4286
= P ( Z <1.4286) From Standard Normal Table
= 0.9234
P(74.5 < X < 76) = 0.9234-0.2375 = 0.6859
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