Question

the weights of bags of baby carrots are normally distributed, with a mean of 31 ounces and a standard deviation of 0.38 ounce. bags in the upper 4.5% are too heavy and must be repackaged. what is the most a bag of baby carrots can weigh and not need to be repackaged? click here to view page 1 of the standard normal table. click here to view page 2 of the standard normal table. a bag of baby carrots can weigh at most ounces without needing to be repackaged. (round to two decimal places as needed.)

Explanation:

Step1: Find the z - score

The upper 4.5% means the area to the right of the z - score is 0.045. So the area to the left of the z - score is $1 - 0.045=0.955$. Looking up in the standard - normal table, the z - score corresponding to an area of 0.955 is approximately 1.70.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $z$ is the z - score, $x$ is the value from the original distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $\mu = 31$, $\sigma=0.38$, and $z = 1.70$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.

Step3: Calculate the value of $x$

Substitute the values into the formula: $x = 31+1.70\times0.38$. First, calculate $1.70\times0.38 = 0.646$. Then, $x=31 + 0.646=31.646$. Rounding to two decimal places, $x\approx31.65$.

Answer:

31.65