Question

A survey indicates that for each trip to a supermarket, a shopper spends an average of 43 minutes with a standard deviation of 12 minutes in the store. The lengths of time spent in the store are normally distributed and are represented by the variable X. A shopper enters the store. Find the probability that the shopper will be in the store for each interval of time listed below. a) Find the probability that the shopper will be in the store between 33 and 66 minutes.b) Find the probability that the shopper will be in the store for more than 39 minutes. Hint: Convert the normal distribution X to Standard normal usingZ formula Z=(X-μ)/σ and then look the Z-values from the table and then find the probability.61-year-old woman doing exercises to prepare for GED in nine days please help I know numbers need to be inserted into the formula I just don’t know how I have not taken math in 30 years please help me in desperate need Please help me step-by-step so I can learn how to do it on my own

Answer 1

We are given that x N(43,12)

To find the probability that the shopper is in the store between 33 and 66 minutes, we will calculate the Z score at each of these times.

Z = 33-43/12 = -0.833

P = 0.336

Z = 66-43/12 = 1.92

P = 0.9726

To find the probability inbetween these values we will minus one from another

P = 0.9726 -0.336

= 0.6366

Answer to part a). P = 0.6366

b) We will calculate the z score at 39 minutes:

Z = (39-43)/12 = -0.33

P(x<39) = 0.3707. The Z table gives us the probability that ZThe probability that the shopper stays in the shop for more that 39 minutes is 1 minus this probability.

1-0.3707 = 0.6293

Answer to part b). = 0.6293