Question

JenStar tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of $16,900.00 and a standard deviation of about $650.00. Using this information, answer the following questions.
For full marks your answer should be accurate to at least three decimal places.
Compute the probability that tomorrow's profit will be

a) between $18,050.50 and $18,603.00

b) less than $17,920.50 or greater than $18,271.50

c) greater than $17,062.50

d) less than $16,835.00 or greater than $18,083.00

e) less than $17,985.50

Answer 1

This is a normal distribution question with




a) P(18050.5 < x < 18603.0)=?

This implies that
P(18050.5 < x < 18603.0) = P(1.77 < z < 2.62) = P(Z < 2.62) - P(Z < 1.77)
P(18050.5 < x < 18603.0) = 0.9956035116518787 - 0.9616364296371288




b) P(X < 17920.5 or X > 18271.5)=?

This implies that
P(X < 17920.5 or X > 18271.5) = P(z < 1.57 or z > 2.11) = 0.959


c) P(x > 17062.5)=?
The z-score at x = 17062.5 is,

z = 0.25
This implies that
P(x > 17062.5) = P(z > 0.25) = 1 - 0.5987063256829237



d) P(X < 16835.0 or X > 18083.0)=?

This implies that
P(X < 16835.0 or X > 18083.0) = P(z < -0.1 or z > 1.82) = 0.4946


e) P(x < 17985.0)=?
The z-score at x = 17985.0 is,

z = 1.6692
This implies that

PS: you have to refer z score table to find the final probabilities.
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