The mean incubation time for a type of fertilized egg kept at 100.4 F is 19 days

Question

The mean incubation time for a type of fertilized egg kept at 100.4 F is 19 days. Suppose that the incubation times are approximatley normally distributed with a standard deviation of 2 days. (a) What is the probability that a randomly selected fertilized egg hatches in less that 15 days? (b) What is the probability that a randomly selected fertilized egg hatches between 17 and 19 days is ? (c) What is the probability that a randomly selected fertilized egg takes over 23 days to hatch is? (d) Would it be unusual for an egg to hatch in less than 13 days? Why? The probability of this event is ____, so it __________ be unusual because the probability is _______ than 0.05. (round to four decimal places as needed)

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    15      
u = mean =    19      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    -2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2   ) =    0.022750132 [ANSWER]

*************

b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    17      
x2 = upper bound =    19      
u = mean =    19      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.341344746   [ANSWER]

*****************

c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    23      
u = mean =    19      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2   ) =    0.022750132 [ANSWER]

*******************

d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    13      
u = mean =    19      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    -3      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -3   ) =    0.001349898 [ANSWER]

Hence,

The probability of this event is [0.001349898], so it [WILL] be unusual because the probability is [LESS] than 0.05. [ANSWER]

********************

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.