In psychology, there is a particular Mental Development Index (MDI) used in the study of infants. The scores on the MDI have approximately a normal distribution with a mean of 100 and standard deviation of 16. We are going to randomly select 64 children and average their MDI scores. What is the probability that the average is greater than 102?

0.1587

Step-by-step explanation:

Here, mean=μ=100 and standard deviation=σ=16.

We have to find  P(average MDI scores of 64 children > 102)=P(xbar>102).

n=64

μxbar=μ=100

σxbar=σ/√n=16/√64=16/8=2

P(xbar>102)=P((xbar-μxbar)/σxbar>(102-100)/2)

P(xbar>102)=P(z>1)

P(xbar>102)=P(0<z<∞)-P(0<z<1)

P(xbar>102)=0.5-0.3413

P(xbar>102)=0.1587

Thus, the probability that the average is greater than 102 is 15.87%

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

600,000,000,000, or 6e14

step-by-step explanation: