Math for Liberal Arts

1. 204 is 2 standard deviations above the mean, so between the mean and 204 would be \(95\% \div 2 = 47.5\%\) of the data.

2. 198 is one standard deviation below the mean. \(68\% \div 2 =34\%\)--so 34% of the data lies between 198 and the mean. \(50\% + 34\% = 84 \%\) of the data lies above 198.

3. 204 is 2 standard deviations of the mean. 2.5% of the data will lie above 204.

   \begin{equation*} 0.025 \left(5,000\right) = 125 \end{equation*}
   There should be 125 bag with more than 204 pieces.

4. 196 is 3 standard deviations below the mean.

   \begin{equation*} 100\% - 99.7\% = 0.3\% \end{equation*}
   \begin{equation*} 0.3\% \div 2 = 0.15\% \end{equation*}
   0.15% of the bag have less than 196 pieces. There is a probability of 0.0015 that a bag has less than 196 pieces.

1. 92.65% is below \(z = 1.45\text{.}\)

   \begin{equation*} 100\% - 92.65\% = 7.35\% \end{equation*}

2. 70.54%

3. 88.49%

4. 83.40% is above \(z = -0.97\text{.}\)

   \begin{equation*} 100\% - 83.40\% = 16.6\% \end{equation*}

5. 88.49% is below \(z = 1.2\) and 99.18% is below \(z = 2.4\text{.}\)

   \begin{equation*} 99.18\% - 88.49\% = 10.69\% \end{equation*}

6. 74.22% is above \(z = -0.65\text{.}\)

   \begin{equation*} 100\% - 74.22\% = 25.78\% \end{equation*}
   72.24% is below \(z = 0.59\text{.}\)
   \begin{equation*} 72.24\% - 25.78\% = 46.46\% \end{equation*}

7. 92.65% is above \(z = -1.45\) and 86.43% is above \(z= -1.1\text{.}\)

   \begin{equation*} 92.65\% - 86.43\% = 6.22\% \end{equation*}

1. \begin{equation*} z = \frac{14 - 21}{4} \end{equation*}
   \begin{equation*} z = \frac{-7}{4} = -1.75 \end{equation*}

2. \begin{equation*} z=\frac{23-21}{4} \end{equation*}
   \begin{equation*} z=\frac{2}{4}=0.5 \end{equation*}

3. \begin{equation*} z=\frac{29-21}{4} \end{equation*}
   \begin{equation*} z = \frac{8}{4}=2 \end{equation*}

1. \begin{equation*} 0.67 = \frac{x - 35}{3} \end{equation*}
   \begin{equation*} 0.67 \cdot 3 = x - 35 \end{equation*}
   \begin{equation*} 2.01 = x - 35 \end{equation*}
   \begin{equation*} x = 37.01 \end{equation*}

2. \begin{equation*} 1.5 = \frac{x - 35}{3} \end{equation*}
   \begin{equation*} 1.5 \cdot 3 = x - 35 \end{equation*}
   \begin{equation*} 4.5 = x - 35 \end{equation*}
   \begin{equation*} x = 39.5 \end{equation*}

3. \begin{equation*} -0.95 = \frac{x - 35}{3} \end{equation*}
   \begin{equation*} -0.95 \cdot 3 = x - 35 \end{equation*}
   \begin{equation*} -2.85 = x - 35 \end{equation*}
   \begin{equation*} x = 32.15 \end{equation*}

First, compute the z-score for \(x = 100\text{,}\) using \(\mu = 120\) and \(\sigma = 12\text{.}\)

Next, use a z-score to table to determine that there a probabilty of 0.9525 that a value will be above that z-score.

The probability that the failure would occur in less than 100 months would be:

1. First, compute z-scores for both \(x = 60\) and \(x = 70\text{.}\)

   \begin{equation*} z = \frac{60 - 67}{10}=\frac{-7}{10}=-0.7 \end{equation*}
   \begin{equation*} z = \frac{70-67}{10}=\frac{3}{10}=0.3 \end{equation*}

   Then look up both z-scores in the table.

   There is a probability of 0.7580 that a value will be above \(z = -0.7\text{.}\)

   \begin{equation*} 1 - 0.7580 = 0.242 \end{equation*}

   There is a probability of 0.242 that a value will be below that z-score.

   There is a probability of 0.6179 that a value will be below \(z = 0.3\text{.}\)

   \begin{equation*} 0.6179 - 0.242 = 0.3759 \end{equation*}

   There is a probability of 0.3759 that the annual rainfall will be between 60 and 70 inches.

2. The z-score corresponding with \(x = 72\) is:

   \begin{equation*} z = \frac{72-67}{10}=\frac{5}{10}=0.5 \end{equation*}

   Using the z-score table, 69.15% of the data lies below \(z = 0.5\text{.}\)

   \begin{equation*} 100\%-69.15\% = 30.85\% \end{equation*}

   30.85% of the data lies above that z-score, or 30.85% of the years studied should have a mean annual rainfall that is more than 72 inches.

   \begin{equation*} 0.3085 \cdot 20 = 6.17 \end{equation*}

   About 6 years.

70% of the finishers will be below the top 30%. \(z = 0.52\) corresponds with 0.6985, which is the probability closest to 0.7 in the table.

Compute the raw score that corresponds with \(z = 0.52\text{:}\)

The top 30% of competitors will have lifted 1110.4 pounds or more.