The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and standard deviation $220. What is the probability that the actual expenses will exceed $1560 in 20 or more weeks during the next year?

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean = np = 52 × 0.3085 = 16.042

The standard deviation =  

The standard deviation =

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

From z tables

P(Y > 20) 0.0903

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

the balance is found at the link

step-by-step explanation: