The slope is going to be the one that has the X by it so your slope is going to be 5x. Then your y intercept is going to be the number behind the slope so the y intercept is positive 1.

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3 years ago

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Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel

laiz [17]

Answer:

0.03 is the probability that for the sample mean IQ score is greater than 103.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

$= 1 - P(z \leq 1.875)$

Calculation the value from standard normal z table, we have,

$P(x > 103) = 1 - 0.970 =0.03= 3\%$

0.03 is the probability that for the sample mean IQ score is greater than 103.

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2 years ago