SL 4.12—Z values, inverse normal to find mean and standard deviation

We are given $\mu =75$, $\sigma =6$, and $x=63$. We need to apply the z-score formula.

$z=\frac{x-\mu }{\sigma }=\frac{63-75}{6}=\frac{-12}{6}=-2$

The z-score is $-2$, meaning $x=63$ lies 2 standard deviations below the mean. The negative sign confirms the value is below the mean.

The correct answer is $z=-2$.

We need the z-score for $x=63$ when $\mu =75$ and $\sigma =6$. Since $63<75$, the z-score must be negative — this immediately eliminates options with positive z-scores.

$z=2$ is positive, but since $x=63$ is below the mean $\mu =75$, the z-score must be negative. Eliminated.

$z=12$ is not only positive but also unreasonably large — $12\times 6=72$ units away from the mean, which is far larger than the actual difference of $75-63=12$. Eliminated.

If $z=-0.5$, then $x=75+(-0.5)(6)=75-3=72\ne 63$. This does not match. Eliminated.

Checking: $\mu +z\sigma =75+(-2)(6)=75-12=63$ ✓. The correct answer is $z=-2$.

We have $\mu =120$, $\sigma =15$, and we want $P(R>138)$.

$z=\frac{138-120}{15}=\frac{18}{15}=1.2$

Since $R=138$ corresponds to $z=1.2$, we have $P(R>138)=P(Z>1.2)$.

The correct answer is $P(R>138)=P(Z>1.2)$.

We need the correct z-score for $x=138$ and since $138>120$ (above the mean), the z-score must be positive.

A negative z-score would correspond to a value below the mean. Since $138>120$, the z-score must be positive. Eliminated.

$z=18$ would mean $138$ is 18 standard deviations above the mean — an absurd result. The actual difference is $138-120=18$ mm, not 18 standard deviations. Eliminated.

$z=0.12$ would mean $x=120+0.12\times 15=121.8$, not $138$. Eliminated.

Confirming: $z=(138-120)/15=1.2$, so $P(R>138)=P(Z>1.2)$. Correct.