Find the probability a shopper, selected at random, spends less than 96 minutes at the mall. Show your work. The correct probability statement is required to get full credit. Write your final answer to 4 decimal places.

The probability a shopper, selected at random, spends less than 96 minutes at the mall is 0.9450

Step-by-step explanation:

Mean = xbar = 88 mins

Standard deviation = σ = 5 mins

To determine the probability the probability a shopper, selected at random, spends less than 96 minutes at the mall, we need to standardize 96 mins, that is, obtain its z-score.

The standardized score is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (96 - 88)/5 = 1.6

To determine the probability the probability a shopper, selected at random, spends less than 96 minutes at the mall, P(x < 96) = P(z < 1.6)

We'll use data from the normal probability table for these probabilities

P(x < 96) = P(z < 1.6) = 1 - P(z ≥ 1.6) = 1 - P(z ≤ -1.6) = 1 - 0.055 = 0.9450

The probability a shopper, selected at random, spends less than 96 minutes at the mall is 0.9450

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.