a. The probability that at least 40 can taste the difference between the two oils is 0.0392.
b. The probability that at most 5% can taste the difference between the two oils is 1.0.
- Let x represent those who can detect a difference in taste.
Given the following data:
- Total population = 100 people.
- No taste difference = 97 people.
- Sample size = 1000 individuals.
How to calculate the probabilities.
First of all, we would determine the probability for the people who can detect a difference in taste between the new and old oils as follows:
People who can = [tex]100 -97[/tex] = 3 people.
[tex]Probability = \frac{3}{100}[/tex]
Probability = 0.03.
Next, we would calculate the mean and standard deviation:
For the mean:
[tex]Mean = nP\\\\Mean = 1000 \times 0.03[/tex]
Mean = 30
For the standard deviation:
[tex]\delta_x = \sqrt{nP(1-P)} \\\\\delta_x = \sqrt{30(1-0.03)}\\\\\delta_x = \sqrt{30\times 0.97}[/tex]
Standard deviation = 5.39.
Note: We would use a continuity correction of 0.5.
a. The probability that at least 40 can taste the difference between the two oils is given by this expression:
[tex]P(x\geq 40)=P(x\geq 40-0.5)\\\\P(x\geq 40)=P(x\geq 39.5)\\\\P(x\geq 40)=1-P(x\leq 39.5)\\\\P(x\geq 40)=1-P(Z\leq \frac{39.5-30}{5.39} )\\\\P(x\geq 40)=1- \phi(1.76)\\\\P(x\geq 40)=1- 0.9608\\\\P(x\geq 40)=0.0392[/tex]
b. The probability that at most 5% can taste the difference between the two oils:
Sample size = [tex]\frac{5}{100} \times 1000 =50\;individuals[/tex]
[tex]P(x\leq 50)=P(x\leq 50+0.5)\\\\P(x\leq 50)=P(x\leq 50.5)\\\\P(x\leq 50)=1-P(Z\leq \frac{50.5-30}{5.39} )\\\\P(x\leq 50)= \phi(3.80)\\\\P(x\leq 50)=1.0[/tex]
Read more on probability here: https://brainly.com/question/25870256
