In: Statistics and Probability
A large food selling store claims that the bags of the brown rice contain an average of 22 kg of rice with the standard deviation of the weights being 0.6 kg. Determine the probability that the mean weight of 10 randomly selected bags of brown rice will be between 21.6 kg and 22.4 kg., if the store’s claim is true. Assume that the weights of bags rice are normally distributed.
0.0174 |
|
0.9652 |
|
0.9826 |
|
0.9131 |
Solution :
Given that,
mean = = 22
standard deviation = = 0.6
n = 10
= 22
= / n = 0.6 / 10
P(21.6 < < 22.4) = P((21.6 - 22) / 0.6 / 10 <( - ) / < (22.4 - 22) / 0.6 / 10))
= P(-2.11 < Z < 2.11)
= P(Z < 2.11) - P(Z < -2.11) Using z table,
= 0.9826 - 0.0174
= 0.9652
Probability = 0.9652