Find $\alpha $ for this dice rolling game.

There is the point A(10) on the number line. Let's following the die rolling game rules like below with its pips are 1 to 6 as commonly we've known.

- The 1st rule) A point moves as much as +2[positive direction with magnitude 2] for the even pips like 2,4 and 6

- The 2nd rule) A point moves as much as −1[negative direction with magnitude 1] for the odd pips like 1,3 and 5.

After doing this trial 100 times, "A" lied on the right direction compared with 90. We get $P(Z\ge \alpha )$ considering the continuity correction. Find the real number $\alpha $. (Z is the random variable following normal distribution.)

From here, my solution starts. Let the ${X}_{i}$ be the movement of a point for ith trial (Here Each i, $1\le i\le 100$)

$$\begin{array}{clcr}{X}_{i}& 2& -1& \\ p({X}_{i})& 1/2& 1/2& 1\end{array}$$

From the table I got $E({X}_{i})=\frac{1}{2}$ and $V({X}_{i})=\frac{9}{4}$

So only have to consider is $X(={X}_{1}+{X}_{2}+...+{X}_{100})>80$, Plus $E(X)=50$ and $V(X)={15}^{2}$

Therefore $P(X\ge 80.5)=P(Z\ge \frac{80.5-50}{15})$ from continuity correction for $P(X>80)$. (In other words, $\alpha =\frac{61}{30}(=\frac{30.5}{15})$ )

The answer sheet claimed the $\alpha =2.1$. I can't find any errors in my solution. Please let me know which points did I wrong.

p.s.) The answer sheet solution. (The reason why the author claimed 2.1).Say Y be the number of the even pips doing the 100 times trials, Then only just find $P(10+2Y-(100-Y)>90)$

In other words $P(Y\ge 61)$. Since the Y~B(100,1/2), $P(Y\ge 61)=P(Y\ge 60.5)=P(z\ge 2.1)$

There is the point A(10) on the number line. Let's following the die rolling game rules like below with its pips are 1 to 6 as commonly we've known.

- The 1st rule) A point moves as much as +2[positive direction with magnitude 2] for the even pips like 2,4 and 6

- The 2nd rule) A point moves as much as −1[negative direction with magnitude 1] for the odd pips like 1,3 and 5.

After doing this trial 100 times, "A" lied on the right direction compared with 90. We get $P(Z\ge \alpha )$ considering the continuity correction. Find the real number $\alpha $. (Z is the random variable following normal distribution.)

From here, my solution starts. Let the ${X}_{i}$ be the movement of a point for ith trial (Here Each i, $1\le i\le 100$)

$$\begin{array}{clcr}{X}_{i}& 2& -1& \\ p({X}_{i})& 1/2& 1/2& 1\end{array}$$

From the table I got $E({X}_{i})=\frac{1}{2}$ and $V({X}_{i})=\frac{9}{4}$

So only have to consider is $X(={X}_{1}+{X}_{2}+...+{X}_{100})>80$, Plus $E(X)=50$ and $V(X)={15}^{2}$

Therefore $P(X\ge 80.5)=P(Z\ge \frac{80.5-50}{15})$ from continuity correction for $P(X>80)$. (In other words, $\alpha =\frac{61}{30}(=\frac{30.5}{15})$ )

The answer sheet claimed the $\alpha =2.1$. I can't find any errors in my solution. Please let me know which points did I wrong.

p.s.) The answer sheet solution. (The reason why the author claimed 2.1).Say Y be the number of the even pips doing the 100 times trials, Then only just find $P(10+2Y-(100-Y)>90)$

In other words $P(Y\ge 61)$. Since the Y~B(100,1/2), $P(Y\ge 61)=P(Y\ge 60.5)=P(z\ge 2.1)$