Assume that the duration of human pregnancies can be described by a Normal distribution with mean 278 days and standard deviation 12 days. About 15% of babies are born prematurely. What is the length of a pregnancy that is the cutoff between being born prematurely and not being born prematurely?

X  =  262 days

Step-by-step explanation:

For Normal Distribution N( 0,1) we have z table and we can find z score for value of 15 % or  0,15. As 0,15 is not in the table we extrapolate then

           0,1492       ⇒  1,4

           0,1515        ⇒  1,3               0,15 ⇒  x     0,1515  - 0,15  = 0,0015

Δ    =    0,0023     ⇒   0,1

By rule of three

If       0,0023      ⇒   0,1

         0,0015     ⇒      x

x  =  0,0652

Therefore the value between  1,3  and  1,4 is:

1,4  -  0,0652  

z (score ) = 1,3348  and as we are looking for values below the mean ( that is at the left of the bell shape curve we need to change the value to a negative one  z = - 1,3348

Now  when  N ( 278 , 12 ) is normalized to  N ( 0 , 1)

z( score ) = ( X - μ ) / σ

- 1,3348 * 12  = X - 278

X = 278 - 16,0176

X  = 278 - 16

X  =  262 days

x>7

Step-by-step explanation:

Because 2 tons of ore in an hour  

It says that the government purchase 4 tons, and 10 tons are sold to the mineral extractor  

4+10=14

14÷2=7