Find value in standard normal distribution equation

Question

I'm given the following equation: $P(X<b)=0.05$ and $X \sim N(-1, 4)$. Now I have to find the value of b in this equation. First, I convert $P(X<b)$ to be using the standard normal distribution. This gives $P(Z<\frac{b - \mu} \sigma) = 0.05$ or $P(Z<\frac{b + 1} 2)=0.05$. By looking into a table with all values for $\phi(z)$, I found out that $\frac{b + 1} 2 = -3.29$ or $b = -7.58$. The solution that was given by the teacher is $-4.29$.

I don't know what I did wrong?

Answer 1

If you look at this table you will seee that $\Phi(-1.645)=0.05$

Thus the equation is $\frac{b+1}{2}=-1.645$

I used linear interpolation. The mean of $\Phi(-1.64)$ and $\Phi(-1.65)$. This is almost $0.05$

Therefore $b=-4.29$

Answer 2

It seems like the way you used the table of $\Phi(z)$ is incorrect. Using this table: We see that $z\approx 1.645$ when $\Phi(z)=0.95$. Therefore, where $\Phi(z)=0.05$, $z\approx -1.645$ since the normal distribution is symmetric.

Using this value gives:

$$\frac{b+1}{2}\approx -1.645$$

Interestingly, you found $\frac{b+1}{2}=-3.29$, which is exactly double of $-1.645$.

Solving for $b$ gives: $$b\approx -4.29$$ Which is the answer given by your teacher.