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The answer is below
Step-by-step explanation:
Given that:
The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.
The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
a) For x < 2:
From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%
b) For x = 2:
For x = 11:
From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337
c) For x = 11:
From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525
d) For x = 2:
For x = 5:
From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845
e) For x = 5:
From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033