A scientist measures the concentration of a chemical solution across five different trials… — SAT

A scientist measures the concentration of a chemical solution across five different trials. The concentrations recorded for the first four trials are $12.1$ grams per liter ( $\text{g/L}$ ), $12.2\text{ g/L}$ , $12.4\text{ g/L}$ , and $12.5\text{ g/L}$ . The concentration for the fifth trial is $x\text{ g/L}$ , where $x < 12.0$ . If the mean of the concentrations from all five trials is equal to their median, what is the value of $x$ ?

Cevap: 11.8

Cevap

11.8

The correct answer is

11.8

. Since

x < 12.0

, sorting the five trial concentrations in ascending order gives

x

12.1

12.2

12.4

, and

12.5

. The median is the middle value of this ordered list, which is

12.2

. The mean is the sum of the five values divided by

5

, which is

\frac{x + 12.1 + 12.2 + 12.4 + 12.5}{5} = \frac{x + 49.2}{5}

. Setting the mean equal to the median gives the equation

\frac{x + 49.2}{5} = 12.2

. Multiplying both sides by

5

yields

x + 49.2 = 61

. Subtracting

49.2

from both sides gives

x = 11.8

Adım Adım Çözüm

Order the data set in ascending order using the condition

x < 12.0

x, 12.1, 12.2, 12.4, 12.5

To find the median, the values must be listed in order. Since

x

is less than

12.0

, it must be the smallest value in the data set.

Identify the median of the five trials.

12.2

For a set of

5

ordered values, the median is the third value.

Write an equation representing the mean of the five trials set equal to the median.

\frac{x + 12.1 + 12.2 + 12.4 + 12.5}{5} = 12.2

The problem states that the mean of the five concentrations is equal to their median.

Solve the equation for

x

x = 11.8

Multiply both sides of the equation by

5

to get

x + 49.2 = 61.0

, then subtract

49.2

from both sides.

Anahtar Kavram

Calculating the mean and identifying the median of a data set, and solving for an unknown variable under inequality constraints.

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru