Question

a) The time of arrival of a person at work has a uniform distribution from 8:40am to 9:20am. On any day, what is the probability that the person will not arrive in the office early given that office starts from 9:00am?

b) Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD value of less than 2.5 standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and sex roughly follow a normal distribution.

1) If a group of athletes have a mean BMD of 1.1 and same standard deviation as young adults, what percent of this group have osteoporosis by the WHO criterion?

2) Women aged 60 to 69 are, of course, not young adults. The mean BMD in this age group is -1.2 (negative two) on the scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis?

c} Customers arrive at an ATM inside a shop at a rate of 10 per hour.

1)What is the probability that the next customer will arrive in 30 minutes?

2) What is the probability that there will be no customers next hour?

Answer 1

(A)-

Uniform distribution formula - x-b/a-b here a-8:40 AM, b-9:20 AM, x-9:00 AM

hence, 20/40- 0.5

(B)

1- Mean of the athletic population is the same as young adults

hence area covered by mean(+-)2.5sigma^2 covers is 98.76

We need to consider the area in negative side of Z curve which will be (100-98.76)/2=0.62%

2- Not able to understand the problem pls elaborate it more

(C)

Its possion distribution problem:

Lambda per hour = 10

Lambda per 30 minutes = 10*0.5=5

(1)- P (next cutomer in 30 minutes) - e^(-lambda*time)*(lambda*time)^n/n!

here lambda= 5, n = 1, time = 1

hence e^(-5)*(5)^1/1= 0.03369

(2) P (no cutomer in 60 minutes) - e^(-lambda*time)*(lambda*time)^n/n!

lambda = 10, time =1, n=0

e^(-10)*(10)^0/0!= 0.0000454