Question

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?

Answer 1

Answer:

a. $0.42465\approx 42.5\%$

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.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{103-100}{16}$

$z=\frac{3}{16}$

$z=0.1875$

$z\approx 0.19$

Now, we need to find $p(z>0.19)$.

$P(z>0.19)=1-)(z<0.19)$

Using normal distribution table, we will get:

$P(z>0.19)=1-0.57535$

$P(z>0.19)=0.42465$

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.

$_nC_r\cdot P^r\cdot (1-P)^{n-r}$

For the given scenario $P=0.42465$, $n=9$ and $r=3$.

$\frac{9!}{3!(9-3)!}\cdot (0.42465)^{3}\cdot (1-0.42465)^{9-3}$

$\frac{9*8*7*6!}{3*2*6!}\cdot (0.42465)^{3}\cdot (0.57535)^{6}$

$3*4*7\cdot 0.076576124894625\cdot 0.0362737708038501$

$84\cdot 0.076576124894625\cdot 0.0362737708038501$

$0.23332720349186\approx 0.23$

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