This is what is known as a "sample size calculation" and is used in clinical trials to determine how much data is needed to proove a practical improvement in patient outcome above the existing standard treatment.

The population distribution is N(56,21^2).

The distribution of the sample mean will be N(56,21^2/n)

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This is what is known as a "sample size calculation" and is used in clinical trials to determine how much data is needed to proove a practical improvement in patient outcome above the existing standard treatment.

The population distribution is N(56,21^2).

The distribution of the sample mean will be N(56,21^2/n)

where n is the size of the sample.

a) We want to choose the sample size so that at least 90% of the distribution of the sample mean is expected to be above 52.

The 10th percentile of a standard Normal distribution is (using lookup tables)

-1.282

Converting this to the 10th percentile on our distribution of the sample mean we get

p(n) = -1.282 x 21/sqrt(n) + 56

We want this value to equal 52 at least so that at least 90% of the distribution is above 52.

Therefore we want n such that

-1.282 x 21/sqrt(n) + 56 >= 52

This implies

sqrt(n) >= (-1.282 x 21)/(-4) = 6.728

Therefore n >= 45.27

So n must be at least 46.

b) Given this value of n, the chance that the sample mean will be less than 50 is

`Phi((50-56)/(21/sqrt(46))) = Phi(-1.938) = 0.02632` x 100%

where `Phi` is the distribution function of the standard Normal distribution.

The chance that the sample mean will be less than 51 is

`Phi((51-56)/(21/sqrt(46))) = Phi(-1.615) = 0.05317` x 100%

Therefore the chance that the sample mean is between 50 and 51 is

5.317 - 2.632 = 2.69%

**a) n needs to be at least 46**

**b) the chance is 2.69%**