The mean incubation time for a type of fertilized egg kept at 100.1100.1°F is 23

Question

The mean incubation time for a type of fertilized egg kept at 100.1100.1°F is 2323 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 11 dayday. (a) What is the probability that a randomly selected fertilized egg hatches in less than 2121 days? (b) What is the probability that a randomly selected fertilized egg takes over 2525 days to hatch? (c) What is the probability that a randomly selected fertilized egg hatches between 2222 and 2323 days? (d) Would it be unusual for an egg to hatch in less than 2020 days? Why? (a) The probability that a randomly selected fertilized egg hatches in less than 2121 days is nothing. (Round to four decimal places as needed.) (b) The probability that a randomly selected fertilized egg takes over 2525 days to hatch is nothing. (Round to four decimal places as needed.) (c) The probability that a randomly selected fertilized egg hatches between 2222 and 2323 days is nothing. (Round to four decimal places as needed.) (d) Would it be unusual for an egg to hatch in less than 2020 days? Why? The probability of this event is nothing, so it would would not be unusual because the probability is greater less than 0.05. (Round to four decimal places as needed.)

Explanation / Answer

Given, population mean, mu=23, population standard deviation, sigma=11. Use, Z score formula to compute the Z score. Next look into Z table to find area corresponding to the Z score.

Z=(X-mu)/sigma, where, X is raw score.

a. P(X<23)=P[Z<(23-23)/11]=P(Z<0)=0.5 (ans)

b. P(Z>25)=P[Z>(25-23)/11]=P(Z>0.18)

=1-0.5714 [The Z table gives area under standard normal curve to the left of Z, therefore, subtract the area from 1 to find area greater than the Z score].

=0.4286 (ans)

c. Find Z scores corresponding to X1=22 and X2=23

Z1=(22-23)/11=-0.09, and Z2=(23-23)/11=0

The two Z scores are of opposite signs, find area between mean and Z scores respectively and add them.

P(22<X<23)=0.0359+0.0000=0.0359 (ans)

d. P(X<20)=P[Z<(20-23)/11]

=P(Z<-0.27)

=0.3936

The probability of the event is greater than 5% that is 0.05, therefore, the event cannot be considered unusual.

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