There are $1000$ students whose average heights reach $174,5$ cm normally distributed with standard deviation of 6,9. How many students that expected to own the height:
a) less than $164$ cm?
b) between $171,5$ cm and $182$ cm?
c) more than or equally $188$ cm?
Attempt: By using z test, we should plug in $\overline{x}, \mu, \sigma$ given in the question above. $z = \dfrac{\overline{x} - \mu} {\sigma}$ but I am having some trouble plugging the value of $\overline{x}$ because in my textbook, it got a correction factor $\pm 0,5$ in that case (depending on lower/upper one). At least can you give me the value of $\overline{x}$ for that three points, and your briefly explanation shall be needed to clear the bush.
Second, what are the differences of "more than" AND "more than or equally" to the value of $\overline{x}$. Are they have a same value?
And yet, we need to find $P(Z > z)$ (for example, question a) by looking at $z$ table, isn't it? Thanks.
The continuity correction factor is necessary for some discrete variables (binomial, poisson) only. But here it is supposed that the heights are normally distributed.
Since we have a continuous random variable here there is no difference. Only for discrete variables we have a difference, statistically. For discrete variables we have $P(X\geq x)=1-P(X\leq x-1)$ and $P(X>x)=1-P(X\leq x)$
Firstly we have to standardize the variable to get a random variable with mean $0$ and variance $1$ in order to use the z-table.
$Z=\frac{X-\mu}{\sigma}$
$P(X\leq x)=\Phi\left(\frac{x-174}{6.9 }\right)$
Plugging in the value for x
$P(X\leq 164)=\Phi\left(\frac{164-174}{6.9 }\right)=\Phi(-1.45)$
Now you can look at a z-table and finde the correponding probability.
If table contains only values for positive z´s then you can use the following relation: Since $Z$ is symmetric around 0 we have $\Phi(-1.45)=1-\Phi(1.45)$