Answer to Question #98803 in Statistics and Probability for Ahmed

Question

the amount of mineral water consumed by a person per day on the job is normally distributed with 19 ounces and standard deviation 5 ounce
a company supplies it's employees with 2000 ounces of mineral water daily .the company has 100 employees

find the probability that the mineral water supplied by company will not satisfy the water demanded by it employees??

suppose that you have a sample of 100 values from a population with mean 500 and with standard deviation 80 .what is the probability that the sample mean will be in the interval (490,510)?

Answer 1

1) $P(\bar X>20)=P(Z>\frac{20-19}{\frac{5}{\sqrt{100}}})=P(Z>2)=0.0228.$ +

2)

$P(490<\bar X<510)=P(\frac{490-500}{\frac{80}{\sqrt{100}}}<Z<\frac{510-500}{\frac{80}{\sqrt{100}}})=\\ =P(-1.25<Z<1.25)=0.7888.$