Respuesta :
Answer:
a) 0% probability that one car chosen at random will have less than 49.5 tons of coal.
b) 0% probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
In this question:
[tex]\mu = 88, \sigma = 0.5[/tex]
a. What is the probability that one car chosen at random will have less than 49.5 tons of coal?
This is the p-value of Z when X = 49.5, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49.5 - 88}{0.5}[/tex]
[tex]Z = -77[/tex]
[tex]Z = -77[/tex] has a p-value of 0.
0% probability that one car chosen at random will have less than 49.5 tons of coal.
b. What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal?
Now [tex]n = 35, s = \frac{0.5}{\sqrt{35}}[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{49.5 - 88}{\frac{0.5}{\sqrt{35}}}[/tex]
[tex]Z = -455.5[/tex]
[tex]Z = -455.5[/tex] has a p-value of 0.
0% probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal
