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
is the sum of nth term (we want to figure this out for first 36 terms)
is the first term (we figured this out to be 9.625)
is the term number (36 for our case)
is the common difference (given as
)