Assume that the sample is taken from a large

Step 1 of 3:

Given the mean \(\mu=120\) millimeters, the standard deviation \(\sigma=5.6\) millimeters.

a). We need to find the probability between 120 and 121.8.

The z-score is the sample mean decreased by the mean divided by the standard deviation.

Then the formula is

\(z=\frac{x-\mu}{\sigma}\)

We know that \(x=120\), \(\mu=120\) and \(\sigma=5.6\)

\(\begin{array}{l} z=\frac{120-120}{5.6} \\ z=\frac{0}{5.6} \\ z=0 \end{array}\)

Hence, z-score is 0.

Then the formula is

\(\mathrm{z}=\frac{x-\mu}{\sigma}\)

We know that \(x=121.8\), \(\mu=120\) and \(\sigma=5.6\)

\(\begin{array}{l} z=\frac{121.8-120}{1.8} \\ z=\frac{1.8}{5.6} \\ z=0.3214 \end{array}\)

Hence, z-score is 0.3214.

Now we have to determine the probability.

\(P(0<z<0.32)=P(z<0.32)-P(z<0)\)

Using area under the normal curve table,

\(P(0<z<0.32)=0.6255-0.5000\)

\(P(0<z<0.32)=0.1255\)

Therefore, the probability that the between 120 and 121.8 is 0.1255.