I am so confused, can someone please help?
The activity of a certain enzyme is measured by counting emissions from a radioactively labeled molecule. For a given tissue specimen, the counts in consecutive 10-second time periods may be regarded (approximately) as repeated independent observations from a normal distribution.
Suppose the mean 10-seconds count for a certain tissue specimen is 1,200 and the standard deviation is 35. For that specimen, let Y represent a 10-second count and let Y (line over this Y) represent the mean of six 10-second counts. Both Y and Y(line above this Y) are unbiased - they each have an average of 1,200 - but that doesn't imply that they are equally good. Find Pr{1,175
I can do the less than/equal to sign but that is what is should be for both Pr's * and Pr{1,175 , and compare the two. Does the comparison indicate that counting for 1 minute and dividing by 6 would tend to give a more precise result than merely counting for a single 10-second time period? How?
If I´m understand it right you want to calculate $P(Y\leq 1175)$, where $Y$ is normal distributed as $Y\sim\mathcal N(1200,35^2)$.
You can transform the random variable $Y$ to $Z=\frac{Y-\mu}{\sigma}=\frac{Y-1200}{35}$. Then $Z$ is standard normal distributed: $Z\sim \mathcal N(0,1)$. Thus we have
$P(X\leq 1175)=\Phi\left(Z\leq \frac{1175-1200}{35} \right)=P\left(Z\leq\frac{-25}{35} \right)=\Phi\left(-5/7 \right)\approx \Phi(-0.71429)$
$\Phi\left(z \right)$ is the cdf of the standard normal distribution. If you use a calculator you´ll get $\Phi(-0.71429)=23.75\%$.
Alternatively you can use a table. This table contains only postitive $z$-values. But this is no problem because the normal distribution is symmetric around the mean. Therefore $\Phi(-0.71429)=1-\Phi(0.71429)$
And $0.71429$ is more or less the average of $0.71$ and $0.72$. Therefore we read off the values $\Phi(0.71)$ and $\Phi(0.72)$ from the table and calculate the mean.
$\Phi(0.71)=0.7611, \Phi(0.72)=0.7642$
$\frac{0.7611+7642}{2}=0.7626$
Thus $\Phi(0.71429)\approx 0.7626$
$\Rightarrow \Phi(-0.71429)=1-\Phi(0.71429)\approx 1-0.7626=0.2374=23.74\%$