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lubasha [3.4K]
3 years ago
11

Assume that adults have IQ scores that are normally distributed with a mean of mu equals μ=105 and a standard deviation sigma eq

uals σ=15. Find the probability that a randomly selected adult has an IQ less than 135.
Mathematics
1 answer:
Over [174]3 years ago
6 0

The probability that a randomly selected adult has an IQ less than

135 is 0.97725

Step-by-step explanation:

Assume that adults have IQ scores that are normally distributed with a mean of mu equals μ = 105 and a standard deviation sigma equals σ = 15

We need to find the probability that a randomly selected adult has an IQ less than 135

For the probability that X < b;

  • Convert b into a z-score using z = (X - μ)/σ, where μ is the mean and σ is the standard deviation
  • Use the normal distribution table of z to find the area to the left of the z-value ⇒ P(X < b)

∵ z = (X - μ)/σ

∵ μ = 105 , σ = 15 and X = 135

∴ z=\frac{135-105}{15}=2

- Use z-table to find the area corresponding to z-score of 2

∵ The area to the left of z-score of 2 = 0.97725

∴ P(X < 136) = 0.97725

The probability that a randomly selected adult has an IQ less than

135 is 0.97725

Learn more:

You can learn more about probability in brainly.com/question/4625002

#LearnwithBrainly

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A theater has a seating capacity of 750 and charges $3 for children, s5 for students, and $7 for adults. At a certain screening
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Answer:

Between 150 and 450

Step-by-step explanation:

We are going to find the number by resolving  a system of linear equations.

First we write the system equations :

C+S+A=750

Where C : children, S : students and A : adults

The equation represents the ''full attendance''

The second equation :

3C+5S+7A=3450

This equation represents the totaled receipts.

The system :

C+S+A=750\\3C+5S+7A=3450

has the following associated matrix :

\left[\begin{array}{cccc}1&1&1&750\\3&5&7&3450\end{array}\right]

By performing elementary matrix operations we find that the matrix is equivalent to

\left[\begin{array}{cccc}1&0&-1&150\\0&1&2&600\\\end{array}\right]

The new system :

C-A=150\\S+2A=600

Working with the equations :

C = 150 + A\\S = 600-2A

Our solution vector is :

\left[\begin{array}{c}C&S&A\end{array}\right] =\left[\begin{array}{c}150+A&600-2A&A\end{array}\right]

For example :

If 0 adults attended ⇒ A = 0

C = 150 + 0 \\C = 150\\S = 600 - 2A\\S = 600

This verify the totaled receipts equation :

150($3)+600($5) = $ 3450

A ≥ 0 ⇒ If A = 0 ⇒ C = 150

C = 150 is the minimum children attendance

From the equation :

S = 600 -2A

S ≥0

600 - 2A ≥ 0

600 ≥ 2A

300≥ A

A is restricted to the interval [ 0, 300]

When A = 0 ⇒ C = 150

When A = 300 ⇒C = 150 + A = 150 + 300 = 450

Children ∈ [ 150,450]

With C being an integer number (including 0)

Also S and A are integer numbers (including 0)

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3 years ago
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