Question

As per the records, the data of subjects with Alzheimer’s disease roughly follow normal distribution with

mean age of 80 years and standard deviation 16 years.

a. What is the probability that a subjects’ age is between 63 and 78 years?

b. What is the probability that a subjects’ age is between 74 and 97?

c. What is the probability that a subjects’ age is less than 80?

Answer 1

Solution :

Given that ,

mean = = 80

standard deviation = = 16

a) P(63 < x < 78) = P[(63-80)/16 ) < (x - ) /  < (78-80) /16 ) ]

= P(-1.06 < z <- 0.125 )

= P(z < -0.13 ) - P(z < -1.06 )

Using standard normal table

= 0.4483 - 0.1446 = 0.3037

Probability = 0.3037

b)

P(74< x < 97) = P[(74 - 80)/16 ) < (x - ) /  < (97 - 80) /16 ) ]

= P(-0.38< z <1.06)

= P(z < 1.06 ) - P(z < -0.38)

Using standard normal table

= 0.8554- 0.352= 0.5035

Probability = 0.5035

c)

P(x < 80) = P[(x - ) / < (80 - 80) /16 ]

= P(z < 0 )

= 0.0000

probability= 0.0000