 # Get help with descriptive statistics problems

Recent questions in Descriptive Statistics airesex2 2022-05-01

### For z score, you are taking the sample value subtracting population mean and dividing it by std deviation. Is that correct so far?Now, the "sample value" is defined by an equation. In my scenario, I have a bunch of stores that are reviewed by their customers. The customer amount varies. One store has 2 customers who give review while other store has 19 people who give reviews. For the 1st store in my example, both customers give 5/5. while the 2nd store customers, avg score given by all 19 customers is 4.Sample value is calculated as score/customer count. So for the 1st store, it will be 5/2 while the other store will be 4/19. If I use that in z score eq. it will mean that first store is doing much better than the last store. I would like to know how to make the score "proportional""normalized""fair" etc...Oh my many gods, I can't believe I ended like that. Thanks for pointing that out. Magdalena Norton 2022-05-01

### Without assuming that the diameters of apple pies are distributed according to the normal distributions, estimated the probability that the mean diameter is larger than 32 cm. The sample standard deviation is estimated to be 2. The sample mean is 28 and the sample size is 100.When I used CLT (because the sample size is >30). I am getting a z score of 20? Is this correct? Jayden Mckay 2022-04-07

### What is a latent variable? Alissa Hutchinson 2022-04-07
### Let ${Y}_{1},{Y}_{2},...,{Y}_{n}$ be random samples from a normal distribution where the mean is 2 and the variance is 4. How large must n be in order that $P\left(1.9\le \overline{Y}\le 2.1\right)\ge 0.99$?My attemp^We are computing for the sample mean of our random sample that was given in the problem. By definition, $z=\frac{\overline{Y}-\mu }{\frac{\sigma }{\sqrt{n}}}$. So rewrite the equation so we can transform the data to make the mean 0 and the standard deviation 1. If I do that, I get,$P\left(\frac{1.9-2}{\frac{2}{\sqrt{n}}}\le z\le \frac{2.1-2}{\frac{2}{\sqrt{n}}}\right)\ge 0.99.$ This means that$P\left(\frac{-0.1}{\frac{2}{\sqrt{n}}}\le z\le \frac{0.1}{\frac{2}{\sqrt{n}}}\right)\ge 0.99.$ This also means that$P\left(-0.05\sqrt{n}\le z\le 0.05\sqrt{n}\right)\ge 0.99.$ I am not really sure what to do after this step. I am trying to use the definition of the normal distribution, however that was too difficult to do.Do you guys know what to do after this step? agrejas0hxpx 2022-04-06