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Question
iq scores are normally distributed with a mean of 100 and a standard deviation of 15. what is the probability that a randomly chosen persons iq score will be between 62 and 106, to the nearest thousandth? statistics calculator
Step1: Calculate z - scores for 62 and 106
The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 100$ (mean), $\sigma = 15$ (standard deviation), and $x$ is the value for which we want to find the z - score.
For $x = 62$:
$z_1=\frac{62 - 100}{15}=\frac{- 38}{15}\approx - 2.533$
For $x = 106$:
$z_2=\frac{106 - 100}{15}=\frac{6}{15}=0.4$
Step2: Find the probabilities corresponding to the z - scores
We use the standard normal distribution table (or a calculator with normal distribution functions).
The probability that $Z The probability that $Z The probability that $-2.533 Substitute the values: $0.6554 - 0.0057 = 0.6497$Step3: Calculate the probability between the two z - scores
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0.650 (rounded to the nearest thousandth)