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20 November, 13:50

The mean incubation time of fertilized eggs is 19 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. (a) Determine the 20th percentile for incubation times. (b) Determine the incubation times that make up the middle 97 % of fertilized eggs.

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  1. 20 November, 14:47
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    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
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