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QuestionFind the percent of 3,408 students at Van Buren High with incomes between \$1,000 and \$2,000, and those under \$800.

Studdy Solution

STEP 1

Assumptions1. The number of students is3,408. The mean income is \$1,7513. The standard deviation of the income is \$4214. The income distribution is normal5. We are asked to find the percentage of students with incomes between \$1,000 and \$,000, and the number of students with incomes less than \$800

STEP 2

We will use the standard normal distribution to solve this problem. The standard normal distribution has a mean of0 and a standard deviation of1. We can convert any normal distribution to the standard normal distribution using the formulaZ=X−μσZ = \frac{X - \mu}{\sigma}where- Z is the standard normal random variable- X is the normal random variable (in this case, the income) - μ\mu is the mean of the normal distribution- σ\sigma is the standard deviation of the normal distribution

STEP 3

First, we will find the Z-scores for \$1,000 and \$2,000.
For \1,0001,000Z1000=1000−1751421Z_{1000} = \frac{1000 -1751}{421}For$2,000For \$2,000Z2000=2000−1751421Z_{2000} = \frac{2000 -1751}{421}$

STEP 4

Calculate the Z-scores.
Z1000=1000−1751421=−1.78Z_{1000} = \frac{1000 -1751}{421} = -1.78Z2000=2000−1751421=0.59Z_{2000} = \frac{2000 -1751}{421} =0.59

STEP 5

Now we will find the percentage of students with incomes between \$1,000 and \$2,000. We can do this by finding the area under the standard normal curve between Z_{1000} and Z_{2000}. This can be found using standard normal distribution tables, or a calculator with a normal distribution function.
(1000<X<2000)=(−1.78<Z<0.59)(1000 < X <2000) =(-1.78 < Z <0.59)

STEP 6

Using a standard normal distribution table or a calculator with a normal distribution function, we find that(−1.78<Z<0.59)=0.72(-1.78 < Z <0.59) =0.72So approximately72% of the students had incomes between \$1,000 and \$2,000.

STEP 7

Next, we will find the Z-score for \$800.
Z800=800−1751421Z_{800} = \frac{800 -1751}{421}

STEP 8

Calculate the Z-score.
Z800=800−1751421=−2.26Z_{800} = \frac{800 -1751}{421} = -2.26

STEP 9

Now we will find the percentage of students with incomes less than \$800. We can do this by finding the area under the standard normal curve to the left of Z_{800}. This can be found using standard normal distribution tables, or a calculator with a normal distribution function.
(X<800)=(Z<−2.26)(X <800) =(Z < -2.26)

STEP 10

Using a standard normal distribution table or a calculator with a normal distribution function, we find that(Z<−2.26)=0.012(Z < -2.26) =0.012So approximately.2% of the students had incomes less than \$800.

STEP 11

To find the number of students with incomes less than \$800, we multiply the percentage by the total number of students.
Number of students=Percentage×Total number of studentsNumber\, of\, students = Percentage \times Total\, number\, of\, students

STEP 12

Plug in the values for the percentage and the total number of students to calculate the number of students.
Number of students=0.012×3408Number\, of\, students =0.012 \times3408

STEP 13

Calculate the number of students.
Number of students=0.012×3408=41Number\, of\, students =0.012 \times3408 =41So, approximately41 students had incomes less than \$800.

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