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%

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step-by-step explanation: