Assume that the distribution of the duration of human pregna

Assume that the distribution of the duration of human pregnancies can be approximated with a normal distribution with a mean of 266 days and a standard deviation of 16 days.

(a) What percentage of pregnancies should last between 270 and 280 days?

(b) Find a value x such that 25% of the pregnancies of a duration that is longer than x days.

(c) We select 10 pregnant women at random. What is the probability that the average duration of these 10 pregnancies will be less than 260 days?

(d) We select 60 pregnant women at random. What is the probability that the average duration of these 60 pregnancies will be less than 260 days?

(e) The normal distribution is usually not a good approximation for the distribution of the duration of human pregnancies. In fact, the distribution is usually skewed to the left. Can you think of something that might explain the left skew of the duration of human pregnancies. Discuss. Hint: An induced labour.

(f) If the duration of a human pregnancy is not normally distributed, does that change your answers to parts (c) and (d)? Explain why or why not in each case.

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    270      
x2 = upper bound =    280      
u = mean =    266      
          
s = standard deviation =    16      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.25      
z2 = upper z score = (x2 - u) / s =    0.875      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.598706326      
P(z < z2) =    0.809213047      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.210506721   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1-0.25 =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    266      
z = the critical z score =    0.67448975      
s = standard deviation =    16      
          
Then          
          
x = critical value =    276.791836   [ANSWER]  
     
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C)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    260      
u = mean =    266      
n = sample size =    10      
s = standard deviation =    16      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.185854123      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.185854123   ) =    0.117839957 [answer]

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D)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    260      
u = mean =    266      
n = sample size =    60      
s = standard deviation =    16      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.90473751      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.90473751   ) =    0.001837806 [answer]

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