assume that adults have iq scores that are normally distributed with a mean of 96 and a standard deviation…

assume that adults have iq scores that are normally distributed with a mean of 96 and a standard deviation of 18. find the probability that a randomly selected adult has an iq greater than 121. (hint: draw a graph.)\nthe probability that a randomly selected adult from this group has an iq greater than 121 is \n(round to four decimal places as needed.)

assume that adults have iq scores that are normally distributed with a mean of 96 and a standard deviation of 18. find the probability that a randomly selected adult has an iq greater than 121. (hint: draw a graph.)\nthe probability that a randomly selected adult from this group has an iq greater than 121 is \n(round to four decimal places as needed.)

Answer

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 121$, $\mu=96$, and $\sigma = 18$. $z=\frac{121 - 96}{18}=\frac{25}{18}\approx1.39$

Step2: Find the probability using the standard normal distribution

We want $P(X>121)$, which is equivalent to $P(Z > 1.39)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. From the standard normal table, $P(Z\leq1.39) = 0.9177$. So $P(Z>1.39)=1 - 0.9177=0.0823$

Answer:

$0.0823$