Question

The lengths of pregnancies are normally distributed with a mean of

270270

days and a standard deviation of

1515

days. a. Find the probability of a pregnancy lasting

309309

days or longer. b. If the length of pregnancy is in the lowest

33​%,

then the baby is premature. Find the length that separates premature babies from those who are not premature.Click to view page 1 of the table.

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a. The probability that a pregnancy will last

309309

days or longer is

nothing.

Answer 1

solution

P(x >309 ) = 1 - P(x<309 )

= 1 - P[(x -) / < (309-270) /15 ]

= 1 - P(z < 2.6)

Using z table

= 1 - 0.9953

= 0.0047

probability= 0.0047

(B)

Using standard normal table,

P(Z < z) = 33%

= P(Z < z) = 0.33  

= P(Z <-0.44 ) = 0.33

z = -0.44 Using standard normal table,

Using z-score formula  

x= z * +

x= -0.44 *15+270

x= 263.40