Question Number 4 Answer

pdf

School

Trent University *

*We aren’t endorsed by this school

Course

5210

Subject

Statistics

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by SuperFogCaribou32

Question 4: Suppose you manage a local grocery store. You have concluded that customer expenditures in your store are normally distributed with a mean of $129.53 and a standard deviation of $35.11. (a) What is the probability that a randomly selected customer will spend between $50 and $75? (b) What is the probability that a randomly selected customer will spend between $150 and $165? (c) What is the probability that a randomly selected customer will spend less than $80 or more than $120? (d) How much does a customer need to spend to fall within the top 5% of all customer expenditures? Answers a: The probability that a randomly selected customer will spend between $50 and $75 = P(50<x<75) = P(x<75) – P(x<50) To find the z values, we use the following formula z = 𝑥𝑥−μ σ Where x = 75, σ = 35.11, μ = 129.53 z = 75−129 . 53 35 . 11 = -1.5531 Using the statistics software, P(z < -1.5531) = P (x < 75) = 0.060197381 Where x =50, σ = 35.11, μ = 129.53 z = 50−129 . 53 35 . 11 = -2.2652 Using the statistics software, P(z < - 2.2652 ) = P (x < 50) = 0.01175123 P(50<x<75) = P(x<75) – P(x<50) = 0.060197381 - 0.01175123 = 0.04845 Therefore, the probability that a randomly selected customer will spend between $50 and $75 is 0.04845 or 4.845% approximately. Answers b: The probability that a randomly selected customer will spend between $150 and $165

= P(150<x<165) = P(x<165) – P(x<150) To find the z values, we use the following formula. z = 𝑥𝑥−μ σ Where x = 165, σ = 35.11, μ = 129.53 z = 165−129 . 53 35 . 11 = 1.01025 Using the statistics software, P(z < 1.01025 ) = P (x < 165) = 0.84381307 Where x =150, σ = 35.11, μ = 129.53 z = 150−129 . 53 35 . 11 = 0.5830 Using the statistics software, P(z < 0.5830 ) = P (x < 150) = 0.72006169 P(150<x<165) = P(x<165) – P(x<150) = 0.84381307 - 0.72006169 = 0.12375 Therefore, the probability that a randomly selected customer will spend between $150 and $165 is 0.12375 or 12.38% approximately. Answers c: The probability that a randomly selected customer will spend less than $80 or more than $120? P(x<80) + P(x>120) = P(x<80) + (1 – P(x<120)) To find the z values, we use the following formula. z = 𝑥𝑥−μ σ Where x = 80, σ = 35.11, μ = 129.53 z = 80−129 . 53 35 . 11 = -1.41071 Using the statistics software, P(z < -1.41071 ) = P (x < 80) = 0.07916519 Where x =120, σ = 35.11, μ = 129.53

z = 120−129 . 53 35 . 11 = -0.2714 Using the statistics software, P(z < -0.2714 ) = P (x < 120) = 0.39302915 P(x>120) = 1 – P(x<120) = 1 - 0.39302915 = 0.60697085 P(x<80) + P(x>120) = 0.07916519 + 0.60697085 = 0.686135 Therefore, the probability that a randomly selected customer will spend less than $80 or more than $120 is 0.686135 or 68.61% approximately. Answers d: Using the statistics software, the corresponding z value for the top 5% is 1.64 As P(z >1.64) = 1 - P(z <1.64) = 0.05050258 or 5% approximately. To find x, we use the equation x= μ + z σ , where μ = 129.53, z = 1.64 and σ = 35.11 x = 187.11 Therefore, a customer needs to spend above $187.11 to fall within the top 5% of all customer expenditures.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Question Number 4 Answer

Related Documents