Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean = 246 days and standard deviation 13 days. Complete parts (a) through (1) below

(a) What is the probability that a randomly selected pregnancy lasts less than 242 days?
The probability that a randomly selected pregnancy lasts less than 242 days is approximately (Round to four decimal places as needed)
a

Answer 1

0.3783

Step-by-step explanation

The lengths of the pregnancies, X, is normally distributed with a mean of 246 days and a standard deviation of 13 days,

`X=N(246,13)`

We have to find the probability that a randomly selected pregnancy lasts less than 242 days,

`P(X<242)=?`

For this, we have to standardize the variable X with the formula,

`Z=(X-\mu)/(\sigma)`

So the probability we have to solve is,

`P(X<242)=P\mleft((X-\mu)/(\sigma)<(242-\mu)/(\sigma)\mright)=P\mleft(Z<(242-246)/(13)\mright)=P(Z<-0.31)`

This is equivalent to,

`P(Z<-0.31)=P(Z>0.31)`

Which is also equivalent to,

`P(Z>0.31)=1-P(Z<0.31)`

We have to find these equivalences because, usually, normal distribution tables show the probabilities for positive z-scores and to the left of those values - i.e. less than those values. Find z = 0.31 in a z-table,

So, the probability is,

`P(X<242)=1-P(Z<0.31)=1-0.6217=0.3783`

Hence, the probability that a randomly selected pregnancy lasts less than 242 days is 0.3783.