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

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