Question

The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. Suppose 50 women who recently bore children in a U.S. hospital are randomly surveyed. Assume length of stay is normally distributed.

8. In words, X =

9. X ~ _____(_____,_____)

10. In words X-bar is:

11. _____ ~ _____(_________,________) *(round sd
to 2 decimal places)*

12. Find the probability that a randomly selected woman stays in the hospital more than 3 days.

13. Find the probability that average stay of 50 women is more than 3 days.

14. Which is more likely: (and state why)

A. An individual woman stayed less than two days.

B. The average stay of 50 women was less than two days.

Answer #1

Solution 8:

X = the length of a maternity stay in a U.S. hospital, in days

Solution 9:

X ~ N (2.4, 0.9)

Solution 10:

X-bar = the average length of a maternity stay in a U.S. hospital, in days

Solution 11:

X-bar ~ N (2.4, 0.9/ = 0.13) or N (2.4, 0.13)

Solution 12:

The respective Z-score with X = 3 is

Z =

Z = (3 - 2.4)/0.9

Z = 0.67

Using Z-tables, the probability is

P [Z > 0.67] = 1 - P (Z 0.67)

= 1 - 0.7486

= 0.2514

Solution 13:

The respective Z-score with X-bar = 3 is

Z =

Z = (3 - 2.4)/0.13

Z = 4.71

Using Z-tables, the probability is

P [Z > 4.71] = 0.000

Solution 14:

No because the probability is zero

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