Understanding Z Scores with Practical Examples

A die is rolled 60 times.
Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times.
The exact chance that the face with six spots appears 9, 10, or 11 times?
my solution; in one roll of the die $P(6\text{ spots})=\frac{1}{6}$ In 60 rolls of the die the expected number of times the face with 6 spots appears is given by $E(X)=\text{Mean}=\frac{60}{6}=10$ times.
$SE(6spots)=SD=SQR(\frac{60}{6}\cdot <!--\; \cdot \; -->\frac{5}{6})=2.8868$
Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10
$\text{Standard deviation}=2.8868$
Using the continuity correction of 0.5 the following z-scores are found: $z_1=(9.5-10)/2.8868=-0.1732$ and $z_2=(10.5-10)/2.8868=0.1732$ Subtracting the cumulative probabilities for the two z-scores gives $0.5688-<!--\; -\; -->0.4312=0.1376$ which is the approximate chance that the face with six spots appears 10 times. find the six spots appears between 9,10, or 11?
The exact chance that the face with six spots appears 10 times is found from
$60C10{\left(\frac{1}{6}\right)}^{10\times }{\left(\frac{5}{6}\right)}^{50}=0.137$ find six spots appears 9, 10 or 11 times?

What is the unit of a z-score?

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The​ z-scores are?

I have two cases:
1. If I add a constant to them, does the mean, standard deviation and z score change?
What I think: Mean changes, std deviation doesn't change and z score doesn't change.
2. If I blackuce all the numbers by 5% does the mean, standard deviation and z score change?
What I think: Mean changes by 5%, standard deviation changes but the percentage difference remains the same, and the z score changes by 2.5%
Am I thinking correctly?

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What is the z-score of sample X, if $n=256,\mathrm{\mu <!--\; \mu \; -->}=88,\text{St.Dev}=128,{\mathrm{\mu <!--\; \mu \; -->}}_X=96$?

A sample of n = 20 scores is transformed into z-scores. What is the mean for the set of 20 z-scores?
Select one:
a. 1
b. 10
c. 0
d. cannot be determined without more information

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What is the z-score of X, if n = 72, ${\displaystyle \mu =137}$, SD =25, and X =13?

What is the z-score of sample X, if $n=55,\mathrm{\mu <!--\; \mu \; -->}=38,\text{St.Dev}=66,{\mathrm{\mu <!--\; \mu \; -->}}_X=43$?

People with z-scores above 2.5 on an IQ test are sometimes classified as geniuses. If IQ scores have a mean of 100 and a standard deviation of 16 points, what IQ score do you need to be consideblack a genius?

A set of data has a normal distribution with a mean of 180 and a standard deviation of 20. What percent of the data is in the interval 140 - 220?

What is the z-score of sample X, if ${\displaystyle n=100,\mu =61,St.Dev.=43,\text{and}E\left[X\right]=98}$?

How do you find z-score given the data:-6, -9, -6, -4, 10? 1. 0.2321 2. -1.737 3. 0.8685 4. 0.9355 5. 6.500?

If $X$ is $N(\mathrm{\mu <!--\; \mu \; -->};{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$, then show that $Z=\frac{(X-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})}{\mathrm{\sigma <!--\; \sigma \; -->}}$ is a standard normal random variable; that is, $N(0;1)$

Here is how far I have gotten:
$P(Z\le <!--\; \le \; -->z)=P(X\le <!--\; \le \; -->z\mathrm{\sigma <!--\; \sigma \; -->}+\mathrm{\mu <!--\; \mu \; -->})={\int <!--\; \int \; -->}_{\frac{x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}}^z\frac{e^{\frac{-<!--\; -\; -->y^2}{2}}}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}\mathrm{d}y.$
I used $x=y\mathrm{\sigma <!--\; \sigma \; -->}+\mathrm{\mu <!--\; \mu \; -->}$ as my change of variable.
1.) Is my result correct so far?
2.) And what is my next step? This integral is out of my scope.

If Z denotes the standard normal random variable , then
$P(-<!--\; -\; -->.58<Z<.23)$?