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

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