The number of column inches of classified advertisements appearing on mondays in a certain daily newspaper is normally distributed with population mean of 320 and population standard deviation of 20 inches. referring to scenario 6-1, for a randomly chosen monday, what is the probability there will be less than 340 column inches of classified advertisement? question 5 options:

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 ) =

= 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

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2