Question

question
iq scores are normally distributed with a mean of 100 and a standard deviation of

1. what percentage of people have an iq score between 81 and 127, to the nearest tenth?

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answer attempt 1 out of 2
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Explanation:

Step1: Identify the distribution and parameters

IQ scores are normally distributed with mean $\mu = 100$ and standard deviation $\sigma$ (assuming $\sigma = 15$ as a common IQ standard deviation, since it's not fully visible but typical). We need to find $P(81 < X < 127)$.

Step2: Calculate z - scores

The z - score formula is $z=\frac{X-\mu}{\sigma}$.
For $X = 81$: $z_1=\frac{81 - 100}{15}=\frac{- 19}{15}\approx - 1.27$
For $X = 127$: $z_2=\frac{127 - 100}{15}=\frac{27}{15}=1.8$

Step3: Find the area between z - scores

We need to find $P(-1.27 < Z < 1.8)$.
Using the standard normal table:
$P(Z < 1.8)=0.9641$
$P(Z < - 1.27)=0.1020$
Then $P(-1.27 < Z < 1.8)=P(Z < 1.8)-P(Z < - 1.27)=0.9641 - 0.1020 = 0.8621$
Converting to percentage: $0.8621\times100 = 86.2\%$ (to the nearest tenth)

Answer:

86.2%