the number of pieces of cat food in a one - cup scoop is approximately normally distributed with a mean of…

the number of pieces of cat food in a one - cup scoop is approximately normally distributed with a mean of 344 pieces and a standard deviation of 16 pieces. if a random sample of 28 scoops of cat food is selected, what is the probability that the mean number of pieces will be more than 350 pieces?\n0.0236\n0.3538\n0.6462\n0.9764
Answer
Explanation:
Step1: Calculate the standard error
The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 16$ is the population standard - deviation and $n = 28$ is the sample size. So, $\sigma_{\bar{x}}=\frac{16}{\sqrt{28}}\approx\frac{16}{5.2915}\approx3.024$.
Step2: Calculate the z - score
The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=350$, $\mu = 344$, and $\sigma_{\bar{x}}\approx3.024$. So, $z=\frac{350 - 344}{3.024}=\frac{6}{3.024}\approx1.98$.
Step3: Find the probability
We want to find $P(\bar{X}>350)$, which is equivalent to $P(Z > 1.98)$ using the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq1.98)$ in the standard - normal table, we find $P(Z\leq1.98)=0.9761$. Then $P(Z > 1.98)=1 - 0.9761 = 0.0239\approx0.0236$.
Answer:
0.0236