Question

Kerri takes 3 vitamin pills per day. Each pill has a defined amount of Vitamin C, but the amount actually varies. Let F be the amount of Vitamin C in a Fab pill in mg. Let I be the amount of Vitamin C in an Iab pill in mg. Let H be the amount of Vitamin C in a Hab pill in mg. We will look at F, I and H. F has a mean of

μ_F

= 31.4 mg with a standard deviation of

σ_F

= 6.9 mg. I has a mean of

μ_I

= 17.8 mg with a standard deviation of

σ_I

=3.9 mg. H has a mean of

μ_H

= 42.8 mg with a standard deviation of

σ_H

= 9.0 mg. The total amount of Vitamin C, T, that Kerri gets from these pills equals T = F + I + H. Assume that F, I and H are independently normally distributed.

	Pill	Fab(F)	Iab(I)	Hab(H)
	Mean	31.4	17.8	42.8
	Standard Deviation	6.9	3.9	9.0
	Measure	mg	mg	mg

g) What is the probability that T is < 90?

h) What is the probability that 90 < T < 100?

i) What is the 87th percentile of the distribution of T?

j) The FDA has a daily limit of 105 mg of Vitamin C from pills. What is the probability Kerri exceeds her daily limit on a randomly selected day?

k) Amounts of Vitamin C intake from day to day are independent of one another. What is the probability that Kerri does not exceed her limit on any day of a five-day work week?

Answer 1

mean of T = 31.4 + 17.8 + 42.8 = 92

std dev of T = √(6.9²+3.9²+9²) = 11.992

g)

P( X ≤    90   ) = P( (X-µ)/σ ≤ (90-92) /11.99)      
=P(Z ≤   -0.17   ) =   0.43378   (answer)

h)

we need to calculate probability for ,                  
P (   90   < X <   100   )  
=P( (90-92)/11.99 < (X-µ)/σ < (100-92)/11.99)                  
                  
P (    -0.167   < Z <    0.667   )   
= P ( Z <    0.667   ) - P ( Z <   -0.17   ) =   
0.7476   -    0.4338   =    0.3139   (answer)

i)

P(X≤x) =   0.87000  
      
z value at 0.87=   1.1264   (excel formula =NORMSINV(0.87))
z=(x-µ)/σ      
so, X=zσ+µ=   1.126   *11.99+92
X =   105.508   (answer)

j)

P ( X ≥   105   ) = P( (X-µ)/σ ≥ (105-92) / 11.99)              
= P(Z ≥   1.08   ) = P( Z <   -1.084   ) =    0.1392   (answer)

k)

P(not exceed) = 1-0.1392 = 0.8608

P(not exceed at any of five day) = 0.8608^5 = 0.4727 (ans)