The length of human pregnancies from conception to birth varies according to a d
ID: 3126542 • Letter: T
Question
The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 269 days and standard deviation 18 days.
(a) What proportion of pregnancies last less than 270 days (about 9 months)?
(Please use 4 decimal places.)
(b) What proportion of pregnancies last between 240 and 270 days (roughly between 8 months and 9 months)?
(Please use 4 decimal places.)
(c) How long do the longest 20% of pregnancies last?
(Please use 2 decimal places.)
The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75.
(d) What are the quartiles of the standard Normal distribution? (Use 2 decimal places.)
Q1 =
Q3 =
(e) What are the quartiles of the distribution of lengths of human pregnancies? Please use 2 decimal places.
Q1 =
Q3 =
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 270
u = mean = 269
s = standard deviation = 18
Thus,
z = (x - u) / s = 0.055555556
Thus, using a table/technology, the left tailed area of this is
P(z < 0.055555556 ) = 0.522152064 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 240
x2 = upper bound = 270
u = mean = 269
s = standard deviation = 18
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.611111111
z2 = upper z score = (x2 - u) / s = 0.055555556
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.053577754
P(z < z2) = 0.522152064
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.468574311 [ANSWER]
************************
c)
First, we get the z score from the given left tailed area. As
Left tailed area = 1-0.20 = 0.8
Then, using table or technology,
z = 0.841621234
As x = u + z * s,
where
u = mean = 269
z = the critical z score = 0.841621234
s = standard deviation = 18
Then
x = critical value = 284.1491822 [ANSWER]
************************
d)
Q1:
We get the z score from the given left tailed area. As
Left tailed area = 0.25
Then, using table or technology,
Q1 = z = -0.67448975 [ANSWER]
****
Q3:
We get the z score from the given left tailed area. As
Left tailed area = 0.75
Then, using table or technology,
Q3: z = 0.67448975 [ANSWER]
*************************
e)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.25
Then, using table or technology,
z = -0.67448975
As x = u + z * s,
where
u = mean = 269
z = the critical z score = -0.67448975
s = standard deviation = 18
Then
Q1 = x = critical value = 256.8591845 [ANSWER]
**************
Q3:
First, we get the z score from the given left tailed area. As
Left tailed area = 0.75
Then, using table or technology,
z = 0.67448975
As x = u + z * s,
where
u = mean = 269
z = the critical z score = 0.67448975
s = standard deviation = 18
Then
x = critical value = 281.1408155 [ANSWER, Q3]
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