Answer:
The proportion of bricks that crack during the firing process is 0.0001 = 0.01%.
Step-by-step explanation:
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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Mean of 925 and a standard deviation of 60 degrees.
This means that [tex]\mu = 925, \sigma = 60[/tex]
The proportion of bricks that crack during the firing process is closest to
This is 1 subtracted by the pvalue of Z when X = 1150. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1150 - 925}{60}[/tex]
[tex]Z = 3.75[/tex]
[tex]Z = 3.75[/tex] has a pvalue of 0.9999
1 - 0.9999 = 0.0001
The proportion of bricks that crack during the firing process is 0.0001 = 0.01%.