Using the normal distribution, it is found that 55.7% of response times are between 4 and 6 minutes.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
For this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 4.5, \sigma = 1.2[/tex]
The proportion between 4 and 6 minutes is the p-value of Z when X = 6 subtracted by the p-value of Z when X = 4, hence:
X = 6:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (6 - 4.5)/1.2
Z = 1.25
Z = 1.25 has a p-value of 0.8944.
X = 4:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (4 - 4.5)/1.2
Z = -0.42
Z = -0.42 has a p-value of 0.3372.
0.8944 - 0.3372 = 0.5572 = 55.72%.
Hence:
55.7% of response times are between 4 and 6 minutes.
More can be learned about the normal distribution at https://brainly.com/question/15181104
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