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3 years ago

Suppose IQ scores are normally distributed with mean 100 and standard deviation 10. Which of the following is false?

Mathematics

1 answer:

Trava [24]3 years ago

7 0

93Answer:

C. An IQ score of 80 is more unusual than an IQ score of 120

is the false answer

Step-by-step explanation:

Firstly we need to find the probability of the test score to be less than 90(P<90) then we will continue finding the probability of the IQ score to be between 90 and 110(90<P<110) then we find the probability of the IQ score being more than 110 (P>110).

For P<90

Firstly we compute this using a scientific calculator where we choose the stat option and enter the mean of 100 and a standard deviation of 10, so we check if we make the normal random variable(X) to be 90 the outcome or answer for that is 0.16 so now we know the probability of the IQ score to be less than 90 is 16% chances.

For 90<P<110

then we check with the same method what will be the probability for an IQ score to be between 90 and 110 for a mean of 100 and a standard deviation of 10 we again check a normal random variable(X) of 110 to see what will be the probability of P<110 which we find an answer of 0.84 which is 84% chances so now therefore the probability of an IQ score to be more than 110 is 0.

therefore this tells us an IQ score of 120 is more unusual than an IQ score of 120.

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3 years ago

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We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

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Let us know find its inverse.

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Upon interchanging x and y, we get:

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2 years ago

What is the rate of decay for the function f(x) =2(0.95)^x ( x is on top of the parentheses)

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Answer:

The decay rate is 5%.

Step-by-step explanation:

Let a substance is decaying at the rate of r% per hour from the initial value of P for t hours, then the final value of the substance is given by the function

$f(t) = P(1 - \frac{r}{100})^{t}$ ........... (1)

Comparing this equation with the original equation given as

$f(x) = 2(0.95)^{x}$ ............ (2) we get,

$(1 - \frac{r}{100}) = 0.95$

⇒ $\frac{r}{100} = 0.05$

⇒ r = 5%.

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