00:01
Hello students, as per the given question, let us solve the problem step by step.
00:07
So for the first bit, since the actual weight of the candy bar follows a normal distribution, the sampling distributions of the sample mean weight will also have a normal distributed according to the central limit theorem.
00:28
So coming to the next bit, where we need to find for the mean of the sampling distributions of the sample mean weights will be equal to the mean of the actual weighted distribution.
00:43
So which is the mean value is equals to 2 .2 ohms, like this is according to the central limit theorem also.
00:55
And coming to the next bit where the standard deviation of the sampling distributions of the sample mean weights is the standard deviation of the actual weight distribution divided by square root of the sample size n.
01:07
So the standard deviation will be 0 .04 divided by under root of 5.
01:20
So this is the standard deviation value.
01:22
When calculated, the sigma value is equals to 0 .01788, which is rounded up to three decimals, gives 0 .018 as the answer...