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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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Answer:

(a) 1825 = 2.25x + (2.25-1)(x -108)

(b) 560 mi/h

Step-by-step explanation:

(a) distance = speed·time

The first plane's speed is x. The distance it travels in 2.25 hours is 2.25x.

The second plane's speed is x-108. It travels only 1.25 hours (since it started an hour later). The distance it travels is then (2.25 -1)(x -108).

The problem statement tells us the total of the distances traveled by the two planes is 1825 miles, so we can write the equation ...

... 1825 = 2.25x + (2.25 -1)(x -108)

(b) Simplifying the equation gives ...

... 1825 = 3.50x -135

To solve this 2-step equation, we add 135, then divide by 3.50.

.. 1960 = 3.50x

... 1960/3.50 = x = 560

The first airplane's speed is 560 mph.

<u>Check</u>

In 2.25 hours, the first plane travels (560 mi/h)·(2.25 h) = 1260 mi.

In 1.25 hours, the second plane travels (452 mi/h)·(1.25 h) = 565 mi.

Then 2.25 hours after the first plane leaves, the planes are 1260 +565 = 1825 miles apart, as given in the problem statement.

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