Find the probability a shopper, selected at random, spends less than 96 minutes

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

Assuming the question is "how much more money needs to be raised..."

Add the contributions:

5,358 + 2,834 + 4,132 = 12,324 < this is the current total amount raised. subtract this from the goal, 14,000 to find out much more money is needed for the playground.

14,000 - 12,324 = 1,676

 $1,676  is still needed in order to create the playground.