Answer:
609
Step-by-step explanation:
Standard deviation =
= $30
Margin of error = E = $2
Confidence level = 90%
Since the distribution is said to be normal, we will use z scores to solve this problem.
The z score for 90% confidence level = z = 1.645
Sample size= n = ?
The formula to calculate the margin of error is:
![E=z\frac{\sigma}{\sqrt{n}}\\\\\sqrt{n}=z\frac{\sigma}{E}\\\\n=(\frac{z\sigma}{E} )^{2}](https://tex.z-dn.net/?f=E%3Dz%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%5Csqrt%7Bn%7D%3Dz%5Cfrac%7B%5Csigma%7D%7BE%7D%5C%5C%5C%5Cn%3D%28%5Cfrac%7Bz%5Csigma%7D%7BE%7D%20%29%5E%7B2%7D)
Using the values in above equation, we get:
![n=(\frac{1.645 \times 30}{2} )^{2}\\\\ n = 608.9](https://tex.z-dn.net/?f=n%3D%28%5Cfrac%7B1.645%20%5Ctimes%2030%7D%7B2%7D%20%29%5E%7B2%7D%5C%5C%5C%5C%20n%20%3D%20608.9)
This means, the minimum number of observations required is 609