The question is: the iqs of 9 randomly selected people are recorded. let y(bar) denote their average.

assuming the distribution from which the yi's were drawn is normal with a mean of 100 and a standard deviation of 16, what is the probability that y(bar) will exceed 103?

what is the probability that any arbitrary yi will exceed 103?

what is the probability that exactly 3 of the yi's will exceed 103?

a.

b. 0.23

Step-by-step explanation:

We have been given that the the distribution from which the Yi's were drawn is normal with a mean of 100 and a standard deviation of 16.

(a) We are asked to find the probability that any arbitrary Yi will exceed 103.

Let us calculate z-score corresponding to 103 using z-score formula.

Now, we need to find .

Using normal distribution table, we will get:

Therefore, the probability, that any arbitrary Yi will exceed 103, is 0.42465 or 42.5%.

(b) To find the probability that exactly 3 of the Yi's will exceed 103, we will use binomial probability formula.

For the given scenario , and .

Therefore, the probability, that exactly 3 of the Yi's will exceed 103, is approximately 0.23 or 23%.

answer: 10 turtles

step-by-step explanation: as turtles have four legs so 4.10=40 therefore birds are 4 which sue counted

probably the last option

step-by-step explanation: