Given Information:

\(\displaystyle\mu={133.5}\) and \(\displaystyle\sigma={5.2}\)

\(\displaystyle{n}={230}\)

To find the probability that a single randomly selected value is between 133.6 and 134.1:

z-score is a measure which tells how many standard deviations away a data value is from the mean and is given by the formula:

\(\displaystyle{z}={\frac{{{X}-\mu}}{{\sigma}}}\)

Required probability can be obtained as follows:

\(\displaystyle{P}{\left({133.6}{<}{X}{<}{134.1}\right)}={P}{\left({X}{<}{134.1}\right)}-{P}{\left({X}{<}{133.6}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{134.1}-{133.5}}}{{{5.2}}}}\right)}-{P}{\left({\frac{{{133.6}-{133.5}}}{{{5.2}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{0.12}\right)}-{P}{\left({Z}{<}{0.02}\right)}\)</span>

\(\displaystyle={0.54776}−{0.50798}={0.03978}{\left[{u}{\sin{{g}}}{s}{\tan{{d}}}{a}{r}{d}{\left\|{a}\right\|}{l}{t}{a}{b}\le\right]}\)

Using a standard normal table, look up for z-score 0.12 i.e., 0.1 in the row and 0.02 along the column. The intersection of both values is 0.54776.

Similarly, probability corresponding to z-score 0.02 is 0.50798.

Then, subtract the lower probability value from the higher value.

Therefore, probability that a single randomly selected value is between 133.6 and 134.1 is 0.0398.

\(\displaystyle\mu={133.5}\) and \(\displaystyle\sigma={5.2}\)

\(\displaystyle{n}={230}\)

To find the probability that a single randomly selected value is between 133.6 and 134.1:

z-score is a measure which tells how many standard deviations away a data value is from the mean and is given by the formula:

\(\displaystyle{z}={\frac{{{X}-\mu}}{{\sigma}}}\)

Required probability can be obtained as follows:

\(\displaystyle{P}{\left({133.6}{<}{X}{<}{134.1}\right)}={P}{\left({X}{<}{134.1}\right)}-{P}{\left({X}{<}{133.6}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{134.1}-{133.5}}}{{{5.2}}}}\right)}-{P}{\left({\frac{{{133.6}-{133.5}}}{{{5.2}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{0.12}\right)}-{P}{\left({Z}{<}{0.02}\right)}\)</span>

\(\displaystyle={0.54776}−{0.50798}={0.03978}{\left[{u}{\sin{{g}}}{s}{\tan{{d}}}{a}{r}{d}{\left\|{a}\right\|}{l}{t}{a}{b}\le\right]}\)

Using a standard normal table, look up for z-score 0.12 i.e., 0.1 in the row and 0.02 along the column. The intersection of both values is 0.54776.

Similarly, probability corresponding to z-score 0.02 is 0.50798.

Then, subtract the lower probability value from the higher value.

Therefore, probability that a single randomly selected value is between 133.6 and 134.1 is 0.0398.