Given a normal distribution with a mean of 125 and a standard deviation of 14, what percentage of values is within the interval 111 to 139? (4 points)
32% 

50% 

68% 

95% 

99.7% 
A normal population is given with a mean of 125 and a standard deviation of 14.
To find what percentage of values is within the interval 111 to 139.
Let, X be the random variable.
So, X follows normal with a mean of 125, and standard deviation of 14.
So, (X125)/14~normal(0,1)
To find P(111<X<139)
=P(111125<X125<139125)
=P(14<X125<14)
=P(14/14<(X125)/14<14/14)
=P(1<Z<1)
Where, Z is the standard normal variate.
=phi(1)phi(1)
Where, phi is the distribution function of the standard normal variate.
=phi(1)1+phi(1)
=2*phi(1)1
=2*0.84131
=1.68261
=0.6826.
So, the probability that the values will lie within the interval 111 to 139 is 0.6826.
So, the percentage of values that lie between 111 and 139 is 0.6826*100=68.26%~68% approximately.
So, the answer is (c) 68%
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