A credit card company monitors cardholder transaction habits to detect any unusual activity. Suppose that the dollar value of unusual activity for a customer in a month follows a normal distribution with mean $250 and variance $2400.

(a) What is the probability of $250 to $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654) P-0.4861

(b) What is the probability of more than $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654) P 0.0139

(c) Suppose that 10 customer accounts independently follow the same normal distribution. What is the probability that at least one of these customers exceeds $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654)

Question

Kai Harvey

Mathematics

18 September, 04:30

A credit card company monitors cardholder transaction habits to detect any unusual activity. Suppose that the dollar value of unusual activity for a customer in a month follows a normal distribution with mean $250 and variance $2400.

(a) What is the probability of $250 to $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654) P-0.4861

(b) What is the probability of more than $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654) P 0.0139

(c) Suppose that 10 customer accounts independently follow the same normal distribution. What is the probability that at least one of these customers exceeds $294 in unusual activity in a month? Round your answer to four decimal places (e. g. 98.7654)

+1

Answers (1)

Know the Answer?

Giancarlo Campbell · Answer 1

Step-by-step explanation:

Let x be the random variable representing the dollar value of unusual activity for a customer in a month. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ) / σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 250

σ = √variance = √2400 = 48.99

a) the probability of $250 to $294 in unusual activity in a month is expressed as

P (250 ≤ x ≤ 294)

For x = 250,

z = (250 - 250) / 48.99 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 294

z = (294 - 250) / 48.99 = 0.9

Looking at the normal distribution table, the probability corresponding to the z score is 0.8159

Therefore,

P (250 ≤ x ≤ 294) = 0.8159 - 0.5 = 0.3159

b) the probability of more than $294 in unusual activity in a month is expressed as

P (x > 294) = 1 - P (x < 294)

P (x > 294) = 1 - 0.8159 = 0.1841

c) since n = 10, the formula becomes

z = (x - µ) / (σ/n)

z = (294 - 250) / (48.99/√10) = 2.84

Looking at the normal distribution table, the probability is 0.9977

Therefore, the probability that at least one of these customers exceeds $294 in unusual activity in a month is

1 - 0.9977 = 0.0023