1. ## normal distribution

daily takings at various fast food out lets owned by D&M seem to be normally distributed with mean Rs.5200 and S.D Rs.500
a. what is the probability that the takings from an outlet exceeds Rs.5500
b. what is the probability that the average takings from 12 outlets exceeds Rs.5500

2. Originally Posted by Zandra
daily takings at various fast food out lets owned by D&M seem to be normally distributed with mean Rs.5200 and S.D Rs.500
a. what is the probability that the takings from an outlet exceeds Rs.5500
b. what is the probability that the average takings from 12 outlets exceeds Rs.5500
a. Let X be the random variable takings from an outlet.

Pr(X > 0.5500) = Pr(Z > 0.06) = 1 - Pr(Z < 0.06) = ....

Note: $Z = \frac{X - \mu}{\sigma} = \frac{0.5500 - 0.5200}{0.500} = 0.06$.

b. You need the distribution for $\overline{X}$, the sample mean. From the usual formula that will surely be in your class notes or textbook:

$\overline{X}$ ~ Normal $\left( \mu = 0.5200, \sigma = \frac{0.500}{\sqrt{12}}\right)$.

Now calculate $\Pr(\overline{X} > 0.5500)$.

3. thank you so much