Answer:
Probability that the lifetime of the machine part is less than 13 = 0.6782
Step-by-step explanation:
given that 
Normalizing the function we get





Answer:
Step-by-step explanation:
Since the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = incubation times of fertilized eggs in days
µ = mean incubation time
σ = standard deviation
From the information given,
µ = 19 days
σ = 1 day
a) For the 20th percentile for incubation times, it means that 20% of the incubation times are below or even equal to 19 days(on the left side). We would determine the z score corresponding to 20%(20/100 = 0.2)
Looking at the normal distribution table, the z score corresponding to the probability value is - 0.84
Therefore,
- 0.84 = (x - 19)/1
x = - 0.84 + 19 = 18.16
b) for the incubation times that make up the middle 97% of fertilized eggs, the probability is 97% that the incubation times lie below and above 19 days. Thus, we would determine 2 z values. From the normal distribution table, the two z values corresponding to 0.97 are
1.89 and - 1.89
For z = 1.89,
1.89 = (x - 19)/1
x = 1.89 + 19 = 20.89 days
For z = - 1.89,
- 1.89 = (x - 19)/1
x = - 1.89 + 19 = 17.11 days
the incubation times that make up the middle 97% of fertilized eggs are
17.11 days and 20.89 days
The percentage of the original activity in the sample remains after 17.5 days is obtained as N(t) / No x 100%.
<h3>
What is half life?</h3>
Half-life is the time required for a quantity to reduce to half of its initial value.
The original activity in the sample that remains after 17.5 days is calculated from half life equation as shown below;
N(t) = No(¹/₂)^t/h
where;
- N(t) is the mass of the substance remaining
- No is the initial mass of the substance
- t is time elapsed = 17.5 days
- h is the half life of the substance
percentage of the original activity in the sample remains after 17.5 days;
= N(t) / No x 100%
Thus, the percentage of the original activity in the sample remains after 17.5 days is obtained as N(t) / No x 100%.
Learn more about half life here: brainly.com/question/2320811
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