the total number of forks dropped by customers per day at a busy restaurant is multimodal with a mean of…

the total number of forks dropped by customers per day at a busy restaurant is multimodal with a mean of 24.5 and a standard deviation of 3.3. if a random sample of 80 days is selected, what is the probability that the mean number of forks dropped during those days will be more than 25?\n0.088\n0.152\n0.356\n0.912
Answer
Explanation:
Step1: Identify the formula for z - score
The formula for the z - score of a sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.
Step2: Substitute the given values
We are given that $\mu = 24.5$, $\sigma=3.3$, $n = 80$, and $\bar{x}=25$. $z=\frac{25 - 24.5}{\frac{3.3}{\sqrt{80}}}$ First, calculate $\sqrt{80}\approx8.944$. Then $\frac{3.3}{\sqrt{80}}\approx\frac{3.3}{8.944}\approx0.37$. $z=\frac{25 - 24.5}{0.37}=\frac{0.5}{0.37}\approx1.35$.
Step3: Find the probability
We want to find $P(\bar{X}>25)$, which is equivalent to $P(Z > 1.35)$ using the standard normal distribution. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq1.35) = 0.9115$. So $P(Z > 1.35)=1 - 0.9115=0.0885\approx0.088$.
Answer:
0.088