Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105μ=105 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8989 and 121121.

57.62% probability that a randomly selected adult has an IQ between 89 and 121.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

Find the probability that a randomly selected adult has an IQ between 89 and 121.

This is the pvalue of Z when X = 121 subtracted by the pvalue of Z when X = 89. So

X = 121

has a pvalue of 0.7881

X = 89

has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% probability that a randomly selected adult has an IQ between 89 and 121.

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

the answer is: the length is 5/6 yards.

give me a or a 5 star rating if this is correct!

step-by-step explanation: