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Question

Sixty two percent of 12 th graders attend school in a particular urban school district. If a sample of 500 12 th grade children is selected from the district, find the probability that at least 290 are actually enrolled in that school.

Answer

1.Determine if you can use the normal distribution: n × p = 500 × .62 = 310 n × q = 500 × .38 = 190 These are both larger than 5, so we can use the normal approximation.

2.Find the mean, μ = n × p = 310

3.then µ x q: 310 × .38 = 117.8

4. σ = √(µ x q) = √117.8 = 10.85

5. P (X ≥ 290) using Continuity Correction Factor” = P (X > 289.5)

6. Z-value = (n-μ)/σ = (289.5 – 310) / 10.85 = -1.89

7. The area for -1.819 is .4706 in the z-table

8. Add .5 or substract from .5 depending on which area you are looking for (see image below) => P= .9706, or 97.06%.

2.Find the mean, μ = n × p = 310

3.then µ x q: 310 × .38 = 117.8

4. σ = √(µ x q) = √117.8 = 10.85

5. P (X ≥ 290) using Continuity Correction Factor” = P (X > 289.5)

6. Z-value = (n-μ)/σ = (289.5 – 310) / 10.85 = -1.89

7. The area for -1.819 is .4706 in the z-table

8. Add .5 or substract from .5 depending on which area you are looking for (see image below) => P= .9706, or 97.06%.

Tags

#has-images

Question

Sixty two percent of 12 th graders attend school in a particular urban school district. If a sample of 500 12 th grade children is selected from the district, find the probability that at least 290 are actually enrolled in that school.

Answer

?

Tags

#has-images

Question

Sixty two percent of 12 th graders attend school in a particular urban school district. If a sample of 500 12 th grade children is selected from the district, find the probability that at least 290 are actually enrolled in that school.

Answer

1.Determine if you can use the normal distribution: n × p = 500 × .62 = 310 n × q = 500 × .38 = 190 These are both larger than 5, so we can use the normal approximation.

2.Find the mean, μ = n × p = 310

3.then µ x q: 310 × .38 = 117.8

4. σ = √(µ x q) = √117.8 = 10.85

5. P (X ≥ 290) using Continuity Correction Factor” = P (X > 289.5)

6. Z-value = (n-μ)/σ = (289.5 – 310) / 10.85 = -1.89

7. The area for -1.819 is .4706 in the z-table

8. Add .5 or substract from .5 depending on which area you are looking for (see image below) => P= .9706, or 97.06%.

2.Find the mean, μ = n × p = 310

3.then µ x q: 310 × .38 = 117.8

4. σ = √(µ x q) = √117.8 = 10.85

5. P (X ≥ 290) using Continuity Correction Factor” = P (X > 289.5)

6. Z-value = (n-μ)/σ = (289.5 – 310) / 10.85 = -1.89

7. The area for -1.819 is .4706 in the z-table

8. Add .5 or substract from .5 depending on which area you are looking for (see image below) => P= .9706, or 97.06%.

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#### pdf

owner: naxplast06 - (no access) - Deviant S. - The Practically Cheating Statistics Handbook (2010, CreateSpace Independent Publishing Platform).pdf, p81

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

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