Comparing Sample Mean and a Random VariableLet X ( i ) = ( i = 1 , 2 , … , n + 1 ) be a random sample of size n+1 that is produced from a normal population. Let M be the sample mean of the first n random variables in this random sample.Find the probability that P ( X ( n + 1 ) − M &lt; 0.5 )Variance =1 and Sample Size n=40.I have been stuck in this question for hours since I cannot find a formula to compute X ( n + 1 ) . I tried computing the sample mean including (n+1) and then figure out something between two sample means but I went nowhere.Am I interpreting this question wrong, isn't X ( n + 1 ) just a data value or does that notation mean its the sample mean of n+1 values? I cannot see how this problem could be solved if X ( n + 1 ) is a value ....

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Comparing Sample Mean and a Random VariableLet       X          (      i      )        =  (  i  =  1  ,  2  ,  …  ,  n  +  1  ) be a random sample of size n+1 that is produced from a normal population. Let M be the sample mean of the first n random variables in this random sample.Find the probability that  P  (      X          (      n      +      1      )        −  M  &amp;lt;  0.5  )Variance =1 and Sample Size n=40.I have been stuck in this question for hours since I cannot find a formula to compute       X          (      n      +      1      )      . I tried computing the sample mean including (n+1) and then figure out something between two sample means but I went nowhere.Am I interpreting this question wrong, isn&#039;t       X          (      n      +      1      )       just a data value or does that notation mean its the sample mean of n+1 values? I cannot see how this problem could be solved if       X          (      n      +      1      )       is a value ....

Tyree Duke · Accepted Answer

The random variables       X          n      +      1       and M have the same mean. Let   Y  =      X          +      1        −  M. Then Y has mean 0.The random variable Y is a linear combination of independent normals, so it is normal.Note that       X          n      +      1       has variance 1 and M has variance       1    40  . Thus Y has variance   1  +  (  −  1      )    2        1    40    =      41    40  .Now that we know the mean and the variance of the normally distributed random variable Y, we can find Pr(Y&amp;lt;0.5) in the usual way.

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