Questions: Question 5 3 pts A store is planning on charging 2.99 for a 5lb bag of potatoes. Assume the distribution is approximately normal. If the mean cost of a 5lb bag of potatoes is 3.50 with a standard deviation of 0.25, find the probability of a 5lb bag costing more than the 2.99. This problem is asking for... The cutoff (critical) value ( x ) The probability (area under the curve)

Solution Steps

To find the Z-score for the price of \$2.99, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 2.99 \)
• \( \mu = 3.50 \)
• \( \sigma = 0.25 \)

Substituting the values, we have:

\[ z = \frac{2.99 - 3.50}{0.25} = \frac{-0.51}{0.25} = -2.04 \]

Thus, the Z-score for \$2.99 is:

\[ \text{Z-score for } 2.99: -2.04 \]

Next, we need to find the probability that a 5lb bag of potatoes costs more than \$2.99. This is represented as:

\[ P(X > 2.99) = P(Z > -2.04) \]

Using the cumulative distribution function \( \Phi \), we can express this as:

\[ P(X > 2.99) = \Phi(\infty) - \Phi(-2.04) \]

Since \( \Phi(\infty) = 1 \) and \( \Phi(-2.04) \approx 0.0207 \), we find:

\[ P(X > 2.99) = 1 - 0.0207 = 0.9793 \]

Thus, the probability of a 5lb bag costing more than \$2.99 is:

\[ \text{Probability of a 5lb bag costing more than } 2.99: 0.9793 \]

Final Answer