00:01
Hello student for the given problem we need to find the probability that a randomly selected pregnancy last less than 56 days.
00:10
For that we are going to use the standard normal distribution since we know the mean mu and standard deviation of the population.
00:18
For that first we need to calculate the z -score.
00:21
Formula of it is z equal to x minus mu divided by sigma.
00:25
As we are provided with all the required value our z become 56 minus 160 divided by 12 which come out to be negative 8 .33.
00:37
Now using standard normal distribution table we can find the probability of z less than negative 8 .33 which comes out to be 0.
00:46
So the probability that a randomly selected pregnancy last less than 56 days is approximately 0.
00:53
So from here we can interpret that this is very low probability indicate that extremely rare for a randomly selected, randomly selected pregnancy from this population to last less than 56 days.
01:34
Next in the second part if the hundred pregnant individual was selected independently from this population then we would expect pregnancy to last more than 156 days.
01:48
Here also we will use the z -score which become 156 minus 160 divided by 12 which come out to be negative 0 .33.
01:58
And now we will find the probability that corresponds to the z -score of negative greater than negative 0 .33 and which comes out to be 0 .6306...