B(t)=50(1.4)^t
1 answer:
You might be interested in
Answer:
0.6826 = 68.26% probability that the first machine produces an acceptable cork.
0.933 = 93.3% probability that the second machine produces an acceptable cork.
The second machine is more likely to produce an acceptable cork.
Step-by-step explanation:
When the distribution is normal, we use 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.
What is the probability that the first machine produces an acceptable cork?
The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm, which means that
Acceptable between 2.9 and 3.1, which means that this probability is the pvalue of Z when X = 3.1 subtracted by the pvalue of Z when X = 2.9.
X = 3.1
has a pvalue of 0.8413
X = 2.9
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
0.6826 = 68.26% probability that the first machine produces an acceptable cork.
What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)
For the second machine, we have that . Same probability we have to find out. So
X = 3.1
has a pvalue of 0.9332
X = 2.9
has a pvalue of 0.0002
0.9332 - 0.0002 = 0.933
0.933 = 93.3% probability that the second machine produces an acceptable cork.
Which machine is more likely to produce an acceptable cork?
Second one has a higer probability, so it is more likely to produce an acceptable cork.