In canada, households spent an average of $80.63 cdn monthly for high-speed broadband access.24 assume that the standard deviation is $27.32. if you ask an srs of 500 canadian households with high-speed broadband access how much they pay, what is the probability that the average amount will exceed $85?

Step-by-step explanation:

Assuming a normal distribution for the amount spent by Canadian households for high-speed broadband access, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = amount spent by the Canadian households.

u = mean amount spent monthly.

s = standard deviation

From the information given,

u = $80.63 CDN

s = $27.32 CDN

We want to find the probability that the average amount will exceed $85. It is expressed as

P(x greater than 85) = 1 - P(x lesser than or equal to 85)

For x = 85

z = (85 - 80)/27.32 = 0.18

Looking at the normal distribution table, the corresponding z score is 0.57142

P(x greater than 85) = 1 - 0.57142 = 0.43

i am sure it will be 90 degrees.

step-by-step explanation: