Question

It is known that the average price for a cappuccino in a restaurant is R15.40 with a standard deviation of R1.20. What is the probability that the sampling error for the mean is less than 0.20 for a sample of 31 restaurants?

Answer 1

$$SE=z\times\dfrac{\sigma}{\sqrt{n}}=>z=\dfrac{SE(\sqrt{n})}{\sigma}$$

$$SE<0.2=>z=\dfrac{0.2(\sqrt{31})}{1.2}$$

$$P(SE<0.2)=P(Z<\dfrac{0.2(\sqrt{31})}{1.2})$$

$$\approx P(Z<0.92796)\approx0.823286$$

The probability that the sampling error for the mean is less than 0.20 for a sample of 31 restaurants is $0.823286.$