The incubation time for a breed of chicks is normally distributed with a mean of
ID: 3064622 • Letter: T
Question
The incubation time for a breed of chicks is normally distributed with a mean of 24 days and standard deviation of approximately 1 day. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.)
5. 4 points BBBasicStat8 7.1.008.MI. My Notes Ask Your Teach The incubation time for a breed of chicks is normally distributed with a mean of 24 days and standard deviation of approximately 1 day. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.) Area Under a Normal Curve 2.3 13.5%| 34% 34%|13.5% 35% 68% 95% 99.7% (a) in 22 to 26 days chicks (b) in 23 to 25 days chicks (c) in 24 days or fewer chicks (d) in 21 to 27 days chicks 11:21 AM O Type here to search 3/21/2018Explanation / Answer
a) P(22 < X < 26)
= P((22 - mean)/sd < (X - mean)/sd < (26 - mean)/sd)
= P((22 - 24)/1 < Z < (26 - 24)/1)
= P(-2 < Z < 2)
= P(Z < 2) - P(Z < -2)
= 0.9772 - 0.0228
= 0.9544
The expected mo of chicks = 1000 * 0.9544 = 954.4 = 954
b) P(23 < X < 25)
= P((23 - mean)/sd < (X - mean)/sd < (25 - mean)/sd)
= P((23 - 24)/1 < Z < (25 - 24)/1)
= P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= 0.8413 - 0.1587
= 0.6826
The expected mo of chicks = 1000 * 0.6826 = 682.6 = 683
c) P(X < 24)
= P ((X - mean)/sd < (24 - mean)/sd)
= P(Z < (24 - 24)/1)
= P(Z < 0)
= 0.5000
The expected mo of chicks = 1000 * 0.5 = 500
d) P(21 < X < 27)
= P((21 - mean)/sd < (X - mean)/sd < (27 - mean)/sd)
= P((21 - 24)/1 < Z < (27 - 24)/1)
= P(-3 < Z < 3)
= P(Z < 3) - P(Z < -3)
= 0.9987 - 0.0013
= 0.9974
The expected mo of chicks = 1000 * 0.9974 = 997.4 = 997
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.