Question

x is a normally distributed random variable with mean 97 and standard deviation 7. what is the probability that x is greater than 90? use the 0.68-0.95-0.997 rule and write your answer as a decimal. round to the nearest thousandth if necessary.

Explanation:

Step1: Find the number of standard deviations from the mean

The mean \(\mu = 97\) and standard deviation \(\sigma = 7\). We want to find how many standard deviations 90 is from the mean. Calculate \(z=\frac{90 - \mu}{\sigma}=\frac{90 - 97}{7}=\frac{- 7}{7}=- 1\). So 90 is 1 standard deviation below the mean.

Step2: Apply the 68 - 95 - 997 rule

The 68 - 95 - 997 rule (empirical rule) states that for a normal distribution, about 68% of the data lies within 1 standard deviation of the mean (\(\mu\pm\sigma\)), 95% within 2 standard deviations (\(\mu\pm2\sigma\)), and 99.7% within 3 standard deviations (\(\mu\pm3\sigma\)).

Since the normal distribution is symmetric, the area to the left of \(\mu-\sigma\) (which is 90 here) is \(\frac{1 - 0.68}{2}=0.16\).

We want the probability that \(X>90\), which is the area to the right of 90. So we calculate \(1 - 0.16 = 0.84\).

Answer:

\(0.84\)