Respuesta :
Answer:
0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.
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 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.
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.14 in.
This means that [tex]\mu_{C} = 24, \sigma_{C} = 0.14[/tex]
The width of a door is normally distributed with a mean of 23.87 in and a standard deviation of 0.08 in.
This means that [tex]\mu_{D} = 23.87, \sigma_{D} = 0.08[/tex].
Find the probability that the width of the casing exceeds the width of the door by more than 0.25 in?
This is P(C - D > 0.25).
Distribution C - D:
The mean is:
[tex]\mu = \mu_{C} - \mu_{D} = 24 - 23.87 = 0.13[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{\sigma_{C}^2+\sigma_{D}^2} = \sqrt{0.14^2+0.08^2} = 0.1612[/tex]
Probability:
This probability is 1 subtracted by the pvalue of Z when X = 0.25. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.25 - 0.13}{0.1612}[/tex]
[tex]Z = 0.74[/tex]
[tex]Z = 0.74[/tex] has a pvalue of 0.7704
1 - 0.7704 = 0.2296
0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.
