Question

The average daily jail population in the United States is 706,242. If the distribution is normal and the standard deviation is 52,145, find the probability that on a randomly selected day, the jail population is a. Greater than 750,000 b. Between 600,000 and 700,000

Answer 1

a. The probability that on a randomly selected day, the jail population

is greater than 750,000 is 20.1%

b. The probability that on a randomly selected day, the jail population is

between 600,000 and 700,000 is 43.2%

Step-by-step explanation:

The given is:

1. The average daily jail population in the United States is 706,242

2. The distribution is normal and the standard deviation is 52,145

3. We need to find the probability that on a randomly selected day,

    the jail population is greater than 750,000

4. We need to find the probability that on a randomly selected day,

    the jail population is between 600,000 and 700,000

a.

At first find z-score

∵ z = (x - μ)/σ, where x is the score, μ is the mean and σ is the standard

   deviation

∵  x = 750,000 , μ = 706,242 and σ = 52,145

∴ z = $\frac{750,000-706,242}{52,145}$ ≅ 0.84

Use the normal distribution table of z to find the area to the right of

the z-value

∵ The corresponding area to z-score of 0.84 is 0.79955

- But we are interested in x > 750,000, we need the area to the

  right of z-score

∴ P(x > 750,000) = 1 - 0.79955 = 0.2005

∴ P(x > 750,000) = 0.2005 × 100% = 20.1%

The probability that on a randomly selected day, the jail population is

greater than 750,000 is 20.1%

b.

We will find z-score for 600,000 < x < 700,000

∵ z = $\frac{600,000-706,242}{52,145}$ ≅ -2.04

∵ z = $\frac{700,000-706,242}{52,145}$ ≅ -0.12

Use the normal distribution table of z to find the area between

the two z-values

∵ The corresponding area to z-score of -2.04 is 0.02068

∵ The corresponding area to z-score of -0.12 is 0.45224

- To find P(600,000 < x < 700,000) subtract the two values above

∴ P(600,000 < x < 700,000) = 0.45224 - 0.02068 = 0.4316

∴ P(600,000 < x < 700,000) = 0.4316 × 100% = 43.2%

The probability that on a randomly selected day, the jail population is

between 600,000 and 700,000 is 43.2%

Learn more:

You can learn more about z-score in brainly.com/question/7207785

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