The distribution of the weights of a sample of 140 cargo containers is symmetric and bell-shaped, with a mean of 500 pounds and a standard deviation of 20 pounds. What percentage of the cargo containers will weigh between 460 pounds and 540 pounds? a. 95%
b. Can't tell-there is not enough information
c. 67%
d. 99%

a. 95%

Step-by-step explanation:

We solve this question, using z score formula.

Z score formula = (x - μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation.

n is number of samples

For z1, where x1 = 460, μ = 500, σ = 20, n = 140

z score formula = (460 - 500)/ 20

= -40/20

= -2

We find the probability of the z score using the z score table.

P(x = 460) = P(z = -2)

= 0.02275

For z2, where x2 = 540, μ = 500, σ = 20

z score formula = (540 - 500)/20

= 40/20

= 2

We find the probability of the z score using the z score table.

P(x = 540) = P(z = 2)

= 0.97725

The probability that the cargo containers will weigh between 460 pounds and 540 pounds is calculated as:

= 460 < x < 540

= P(z = 2) - P(z = -2)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage

0.9545 × 100

= 95.45%

Therefore,the percentage of the cargo containers will weigh between 460 pounds and 540 pounds is 95%

215.92 dollars is the answer

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00