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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 incubation times that make up the middle 97% of fertilized eggs is between 17 to 21 days.
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 19 days and a standard deviation of 1 day
a) The 20th percentile correspond to a z score of -2.05
-2.05 = (x - 19)/1
x = 16.95
b) The middle 97% have a z score of ±1.88. Hence:
-1.88 = (x - 19)/1
x = 17.12
1.88 = (x - 19)/1
x = 20.88
The incubation times that make up the middle 97% of fertilized eggs is between 17 to 21 days.
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