Question

1. a cereal company advertises that each box of cereal weighs 16 ounces. in reality, the weights of these boxes follow a normal distribution with a mean of 16.3 ounces and a standard deviation of 0.2 ounce. (a) what proportion of boxes will be above the advertised weight of 16 ounces? (b) find the value of the 25th percentile for this distribution. (c) the weights of “family - size” boxes follow a normal distribution with a standard deviation of 0.3 ounces. if 10% of boxes are less than 32 ounces, what is the mean weight of family - size boxes?

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 16$, $\mu=16.3$, and $\sigma = 0.2$. So, $z=\frac{16 - 16.3}{0.2}=\frac{- 0.3}{0.2}=-1.5$.

Step2: Find proportion for part (a)

We want $P(X>16)$, which is equivalent to $P(Z>-1.5)$ using the standard normal distribution. Since $P(Z > - 1.5)=1 - P(Z\leq - 1.5)$, and from the standard - normal table $P(Z\leq - 1.5)=0.0668$, then $P(Z>-1.5)=1 - 0.0668 = 0.9332$.

Step3: Find z - score for 25th percentile in part (b)

Looking up the z - score in the standard - normal table for the 25th percentile, we find that the z - score $z$ such that $P(Z\leq z)=0.25$ is approximately $z=-0.6745$.

Step4: Calculate value for 25th percentile in part (b)

Using the formula $x=\mu+z\sigma$, with $\mu = 16.3$, $\sigma = 0.2$, and $z=-0.6745$, we get $x=16.3+( - 0.6745)\times0.2=16.3 - 0.1349 = 16.1651$.

Step5: Find z - score for part (c)

We know that $P(X < 32)=0.10$. Looking up the z - score in the standard - normal table for a probability of $0.10$, we find $z\approx - 1.28$.

Step6: Calculate mean for part (c)

Using the formula $z=\frac{x-\mu}{\sigma}$, and re - arranging for $\mu$, we have $\mu=x - z\sigma$. Given $x = 32$, $\sigma = 0.3$, and $z=-1.28$, then $\mu=32-(-1.28)\times0.3=32 + 0.384=32.384$.

Answer:

(a) 0.9332
(b) 16.1651
(c) 32.384