00:01
Let x v length of pregnancies.
00:10
Here given that x follows normal distribution with mean mu equal to 266 days and standard deviation sigma equal to 16.
00:20
Now we have to find for sample of 33, what is the probability that x bar less than 259? that is probability of x bar less than 259.
00:34
First standardize the values, that is z less than x bar minus mu by sigma by root n, that is probability of z less than 259 minus 266 divided by 16 divided by root 33, equal to probability of z less than minus 7 divided by 2 .785, that is probability of z less than minus 2 .51 d2.
01:03
This value obtained using excel formula, non .s .dot s .aist.
01:08
The first parameter is that value and the second parameter is true.
01:13
So probability is 0 .006.
01:20
The probability is 0 .006 into 100 is 0 .6.
01:26
So if 100 samples are taken, then we would expect 0 .6 samples have sample made less than 2509 days.
01:59
In case option 3.
02:04
For e, the null hypothesis is h0 such that mu greater than 266 and h1 such that mu less than are equal to less than 266.
02:16
For that the b value is 0 .006 uptime for b.
02:25
So here the result significant because here 0 .006 is less than 0 .05...