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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 = ≅ 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 = ≅ -2.04
∵ z = ≅ -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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Answer:
C.-2.33
Step-by-step explanation:
We are given that
Significance level, %=0.01
We have to find the critical value of z for a left-tailed test.
To find the critical value , we will use the z-table at significance level 0.01 for left-tailed test.
By using the z-table at significance level 0.01 for left-tailed test
The critical value of z at 1% significance level for left-tailed test
=-2.33
We get p-value by using z-critical value
P-value=0.0099<0.01
So, hypothesis test would be reject the null hypothesis at at 1% significance level.
Hence, the critical value of z=-2.33
Option C is true.