Question

the weights of dimes are approximately normally distributed with a standard - deviation of 0.02 grams. a dime that weighs 2.263 grams is at the 25th percentile. what is the mean weight, in grams, for all dimes? a) 2.255 b) 2.268 c) 2.276 d) 2.513 e) 2.829 although cans of soda are listed as holding 12 ounces, the machine that fills the cans is set to fill the cans with a mean of 12.05 ounces and a standard - deviation of 0.02 ounces. assuming the amount of soda in the cans is normally distributed, which expression represents the amount of soda in a can that is at the 90th percentile of the distribution? a) - 1.282(0.02)+12.05 b) - 0.1(0.02)+12.05 c) 0.1(0.02)+12.05 d) 0.09(0.02)+12.05 e) 1.282(0.02)+12.05

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Find the z - score for the 25th percentile

Looking up the z - score in the standard normal distribution table for the 25th percentile, we find $z\approx - 0.674$. We know $x = 2.263$ grams and $\sigma=0.02$ grams. Substitute into the z - score formula: $-0.674=\frac{2.263-\mu}{0.02}$.

Step3: Solve for the mean $\mu$

Multiply both sides of the equation by $0.02$: $-0.674\times0.02=2.263 - \mu$. Then $- 0.01348=2.263-\mu$. Rearranging gives $\mu=2.263 + 0.01348=2.27648\approx2.276$.

For the second question:

Step1: Find the z - score for the 90th percentile

Looking up the z - score in the standard normal distribution table for the 90th percentile, we find $z\approx1.282$. The mean $\mu = 12.05$ ounces and the standard deviation $\sigma = 0.02$ ounces.

Step2: Use the z - score formula to find the value $x$

Since $z=\frac{x-\mu}{\sigma}$, we can rewrite it as $x=z\sigma+\mu$. Substituting $z = 1.282$, $\sigma=0.02$ and $\mu = 12.05$ gives $x=1.282\times0.02+12.05$.

Answer:

1. C) 2.276
2. E) $1.282(0.02)+12.05$