Question

sat scores in one state is normally distributed with a mean of 1493 and a standard deviation of 176. suppose we randomly pick 36 sat scores from that state.
a) find the probability that one of the scores in the sample is less than 1487.
$p(x < 1487)=$
b) find the probability that the average of the scores for the sample of 36 scores is less than 1487.
$p(\overline{x} < 1487)=$
round each answer to at least 4 decimal places.
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Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(\mu = 1493\), \(\sigma=176\), and \(x = 1487\).

$$z=\frac{1487 - 1493}{176}=\frac{-6}{176}\approx - 0.0341$$

Step2: Find the probability for part (a)

Using the standard normal distribution table (or a calculator with a normal - distribution function, e.g., \(P(X\lt x)=\Phi(z)\) where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution).
\(P(X\lt1487)=\Phi(-0.0341)\approx0.4865\)

Step3: Calculate the z - score for part (b)

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(\mu = 1493\), \(\sigma = 176\), \(n = 36\), and \(\bar{x}=1487\)

$$z=\frac{1487-1493}{\frac{176}{\sqrt{36}}}=\frac{-6}{\frac{176}{6}}=\frac{- 6\times6}{176}=\frac{-36}{176}\approx - 0.2045$$

Step4: Find the probability for part (b)

Using the standard normal distribution table (or a calculator with a normal - distribution function)
\(P(\bar{X}\lt1487)=\Phi(-0.2045)\approx0.4181\)

Answer:

a) \(0.4865\)
b) \(0.4181\)