In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08. Assume that the distribution of percent purity was approximately normal. (a) What percentage of the purity values would you expect to be between 99.5 and 99.7?
Question
In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08. Assume that the distribution of percent purity was approximately normal. (a) What percentage of the purity values would you expect to be between 99.5 and 99.7?
Solution
To solve this problem, we need to use the concept of z-scores in statistics. A z-score measures how many standard deviations an element is from the mean.
Step 1: Calculate the z-scores for 99.5 and 99.7.
The formula for a z-score is:
Z = (X - μ) / σ
where: X is the value we are interested in, μ is the mean, and σ is the standard deviation.
For 99.5: Z = (99.5 - 99.61) / 0.08 = -1.375
For 99.7: Z = (99.7 - 99.61) / 0.08 = 1.125
Step 2: Look up these z-scores in a standard normal distribution table or use a calculator that can calculate the area under the standard normal curve.
The values you get from the table or calculator are the probabilities that a value is less than your given value.
For Z = -1.375, the table gives us 0.0846, which means there is an 8.46% chance a value is less than 99.5.
For Z = 1.125, the table gives us 0.8697, which means there is an 86.97% chance a value is less than 99.7.
Step 3: Subtract the two probabilities to find the percentage of values between 99.5 and 99.7.
0.8697 - 0.0846 = 0.7851
So, we would expect approximately 78.51% of the purity values to be between 99.5 and 99.7.
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