Question

"If the mean IQ is 100 with SD = 15, what is the approximate probability that a randomly chosen sample of 9 people will have a mean IQ greater than 110?"

Answer 1

In this case, the probability that the sample mean of 9 people is greater than 110 in a normally distributed population with a mean of 100 and SD of 15 is approximately 2.28%.

Step-by-step explanation:

The question involves calculating the probability of a sample mean in a normally distributed population. Given that the mean IQ is 100, the standard deviation (SD) is 15, and the sample size (n) is 9, we first need to calculate the standard error (SE) of the mean, which is SD/√n. In this case, SE = 15/√9 = 5.

Next, we convert the sample mean of 110 to a Z-score using the formula: Z = (sample mean - population mean) / SE. So for a sample mean of 110, Z = (110 - 100) / 5 = 10/5 = 2. The Z-score tells us how many standard errors the sample mean is away from the population mean.

To find the probability that the sample mean exceeds 110, we look up the Z-score of 2 in the standard normal distribution table, which gives us the probability to the left of Z. To get the probability to the right (greater than Z), we subtract this from 1. The Z-score of 2 corresponds to a cumulative probability of 0.9772. Therefore, the probability that the sample mean is greater than 110 is 1 - 0.9772 = 0.0228 or 2.28%.

Answer 2

`8.71` %

Step-by-step explanation:

We will find solution to this question by using normal distribution method.

We are required to find

`P(X\geq 110)`

We will find the z-score

`z = (x-\alpha )/(\beta*√(n) )`

Where

`\alpha` is the means and
`\beta` is the standard deviation and n represents the number of sample

Substituting the given values, we get -

`z = (110-100)/(15√(9) ) \\z = 0.223`

Thus, area to the right side of
`0.67` in a normal distribution curve is will represent
`P (Z>0.223)`

`P(X\geq 110)`
`=`
`P (Z>0.223)`

`= 0.0871`