Question

Throwback question: In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood-glucose levels are normally distributed with mean 97.03 mg/dl and standard deviation 29.27 mg/dl. (Note: Round all of your answers to four decimal places where appropriate.)

1. We would expect the blood glucose levels to fall between  and  in 99.7% of patients.
2. A normal glucose range is 65 - 120 mg/dl. What is the probability a patient's blood glucose level falls between these two values?
   • z1 =
   • z2 =
   • probability =
3. What is the probability that a patient's blood glucose level is greater than 107 mg/dl?
   • z =
   • probability =
4. If a blood glucose level is in the top 4%, then the patient is classified as abnormal. What blood glucose level corresponds to this cut-off?
   • z =
   • glucose =  mg/dl

Answer 1

Solution:-

Given that,

mean = = 97.03 mg/dl

standard deviation = = 29.27 mg/dl

a) Using standard normal table,

P( -z < Z < z) = 99.7%

= P(Z < z) - P(Z <-z ) = 0.997

= 2P(Z < z) - 1 = 0.997

= 2P(Z < z) = 1 + 0.997

= P(Z < z) = 1.997 / 2

= P(Z < z) = 0.9985

= P(Z < 2.97) = 0.9985

= z  ± 2.97

Using z-score formula,

x = z * +

x = - 2.97 * 29.27 + 97.03

x = 10.10 mg/dl

Using z-score formula,

x = z * +

x = 2.97 * 29.27 + 97.03

x = 183.96 mg/dl

99.7% fall between 10.10 mg/dl and 183.96 mg/dl

b) P(65 < x < 120) = P[(65 - 97.03)/ 29.27) < (x - ) /  < (120 - 97.03) / 29.27 ) ]

= P(-1.09 < z < 0.78)

= P(z < 0.78) - P(z < -1.09)

Using z table,

= 0.7823 - 0.1379

= 0.6444

c) P(x > 107) = 1 - p( x< 107)

=1- p P[(x - ) / < (107 - 97.03) / 29.27]

=1- P(z < 0.34)

Using z table,

= 1 - 0.6331

= 0.3669

d) Using standard normal table,

P(Z > z) = 4%

= 1 - P(Z < z) = 0.04  

= P(Z < z) = 1 - 0.04

= P(Z < z ) = 0.96

= P(Z < 1.75 ) = 0.96  

z = 1.75

Using z-score formula,

x = z * +

x = 1.75 * 29.27 + 97.03

x = 148.25 mg/dl