Question

At a new exhibit in the museum of science, people are asked to choose between 90 or 120 random draws from a machine. The machine is known to have 93 green balls and 65 red balls. After each draw, the color of the ball is noted and the ball is put back for the next draw. You win a prize if more than 65% of the draws result in a green ballCalculate the probability of getting more than 65% green balls.(Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.) n Probability90 120

Answer 1

Answer:

There is a probability of P₉₀=0.09853 of winning the prize with a 90-balls sample.

There is a probability of P₁₂₀=0.0853 of winning the prize with a 120-balls sample.

Step-by-step explanation:

In this problem, the balls that are removed are then replaced in the total, so the probability of obtaining a green ball is constant and is equal to:

$P_g=\frac{93}{93+65}= 0.5886$

This type of sampling should be analyzed with the binomial distribution, but given the sample size, it is convenient to use the approximation to a normal distribution.

The parameters of the normal distribution are:

$\mu=P_g*n=0.5886*90=52.9746\\\\\sigma=\sqrt{np(1-p)}=\sqrt{90*0.5886*(1-0.5886)} =4.6684$

To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 59 green balls in the sample:

$0.65*90=58.5$

To calculate the probability of getting 59 or more green balls, we have to calculate the z-value

$z=\frac{x-\mu}{\sigma}=\frac{59-52.9746}{4.6684}= 1.29$

Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 59)=P(z\geq 1.29)=0.09853$

There is a probability of P=0.09853 of winning the prize with a 90-balls sample.

When the sample is of 120 balls, we have to recalculate the parameters.

Normal distribution

$\mu=P_g*n=0.5886*120=70.6320\\\\\sigma=\sqrt{np(1-p)}=\sqrt{120*0.5886*(1-0.5886)} =5.3905$

To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 78 green balls in the sample:

$0.65*120=78$

To calculate the probability of getting 78 or more green balls, we have to calculate the z-value

$z=\frac{x-\mu}{\sigma}=\frac{78-70.6320}{5.3905}=1.37$

Then, the probability can be look up with the z-value in a normal distribution table:

$P(X \geq 78 )=P(z\geq 1.37)=0.0853$

There is a probability of P=0.0853 of winning the prize with a 120-balls sample.