In 1992 the American Academy of Pediatrics began recommendin

In 1992, the American Academy of Pediatrics began recommending that healthy infants be placed on their backs when they are put to bed in an attempt to decrease the incidence of sudden infant death syndrome. In 1998, Dr. Beth Davis and her colleagues published a study demonstrating that putting an infant to bed on his or her back did not slow down the pace of infant motor development.

The infants in their sample who were put to bed on their backs began to crawl at an average age of 8.6 motnhs, with a standard deviation od 1.7 months. Asssume that the age at which infants who are to bed on their backs began to crawl is normally distributed. Estimate what percentage of these infants begin to crawl:

Before 10.3 months:________

After 6.9 months: _________

Between 5.2 and 10.3 months:_______

They also found that infants who were put to bed on their stomachs began to crawl at an average age of 7.8 months, with a standard deviation of 2.0 months. Again, assume that these ages are normally distributed.

Before______months, 75% of these infants first crawl.

At_______months, 10% of these infants still have not yet begun to crawl.

The middle 90% of these infants first crawl between ______and______months.

Solution

FIRST PART:

Normal Distribution
Mean ( u ) =8.6
Standard Deviation ( sd )=1.7
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)P(X < 10.3) = (10.3-8.6)/1.7
= 1.7/1.7= 1
= P ( Z <1) From Standard Normal Table
= 0.8413                  
b) P(X > 6.9) = (6.9-8.6)/1.7
= -1.7/1.7 = -1
= P ( Z >-1) From Standard Normal Table
= 0.8413                  
c) To find P(a < = Z < = b) = F(b) - F(a)
P(X < 5.2) = (5.2-8.6)/1.7
= -3.4/1.7 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 10.3) = (10.3-8.6)/1.7
= 1.7/1.7 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(5.2 < X < 10.3) = 0.84134-0.02275 = 0.8186

SECOND PART:                  
a)
P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.674
P( x-u/s.d < x - 7.8/2 ) = 0.75
That is, ( x - 7.8/2 ) = 0.67
--> x = 0.67 * 2 + 7.8 = 9.148                  
b)
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.28
P( x-u/ (s.d) > x - 7.8/2) = 0.1
That is, ( x - 7.8/2) = 1.28
--> x = 1.28 * 2+7.8 = 10.364