Answer:
a) This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.
b) There is a 0.04% .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:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The actual weights of coal loaded into each car are normally distributed, with mean = 50 tons and standard deviation = 0.9 ton. This means that
(a) What is the probability that one car chosen at random will have less than 49.5 tons of coal? (Round your answer to four decimal places.)
This is the pvalue of Z when
has a pvalue of 0.33.
This means that there is a 33% 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? (Round your answer to four decimal places.)
Now we have to use the standard deviation of the sample, since we are working with the sample mean. That is
Now, we find pvalue of Z when
has a pvalue of 0.0004.
This means that there is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal