over the Labor Day weekend would spend an average of $749 (The Associated Press,

Question

over the Labor Day weekend would spend an average of $749 (The Associated Press,
August 12, 2012). Assume that the amount spent is normally distributed with a standard
deviation of $225.
a. What the probability of family expenses for the weekend being less that $400?
b. What is the probability of family expenses for the weekend being $800 or more?
c. What is the probability that family expenses for the weekend will be between $500
and $1000?
d. What would the Labor Day weekend expenses have to be for the 5% of the families
with the most expensive travel plans?

Explanation / Answer

Solution:

Given Mean = 749, Standard deviation = 225
We convert this to standard normal using z = x-/

a) P(X > 400)

= P( X/ > 400749/225)
= P (Z > 1.55)
= 0.9394

b) Given =749 and = 225 we have
P(X >800) =P ( X/ > 800749/225)
= P(Z > 0.23)
= 0.409

c) Given =749 and =225 we have
P(500 < X < 1000 )= P(500749/225< X/ < 1000749/225)
= P(1.11 < Z < 1.12 )
= 0.7351

d) Given =749 and = 225 we have
P(X > x0) = 0.05
=> P(X < x0) = 0.95
=> P(X- / < x0- /) = 0.95
=> P(Z < x0- /) = 0.95
=> x0- / = invNorm(0.95)
=> x0 = + * invNorm(0.95)
Use z-table to get invNorm() = 1.645
=> x0 = 749 +225 *(1.645) = 1119.125
Top 5% are above x0 = 1119.125

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.