The sample mean is μ=320, and sample standard deviation is equal to population standard deviation σₓ=20.
The Z-score is [tex] Z=\frac{ 340-320}{20}=1 [/tex].
Refer to standard normal distribution table.
The required probability is
[tex] P(X<340)=P(Z<1)=0.8413 [/tex]
Let X be the number of column inches of classified advertisements appearing on mondays in a certain daily newspaper.
X follows Normal distribution with mean μ =320 and standard deviation σ=20
The probability that there will be less than 340 column inches of classified advertisement
P(x < 340 ) = [tex] P( \frac{x - mean}{standard deviation} < \frac{340 - 320}{20} ) [/tex]
= P(z < 1)
Using z score table to find probability below z=1 is
P(z< 1) = 0.8413
P(x < 340 ) = P(z<1) = 0.8413
The probability there will be less than 340 column inches of classified advertisement is 0.8413
