Question

Suppose the guests at the hotel have an average weight of 180 with a standard deviation of 50 pounds. An elevator is designed to carry only 10 people with a maximum capacity of 2400 pounds.

5. What is the mean of the sample mean?

6. What is the standard deviation for the sample mean of the 10 people? (Use 2 decimals)

7. If 10 guest ride the elevator at one time, what is the probability that the elevator is overweight? (Hint: If the 10 people weigh more than 2400 pounds, how much does each person have to weigh on average? (Use 4 decimals)

8. If 12 guests ride in the elevator at one time, now what is the probability that the elevator is overweight? (Hint: the standard deviation also changes here) (Use 4 decimals)

Answer 1

Let X be the weight of an individual

X~ Normal ( 180, 50)

Sample size, n = 10

a) Mean of the sample mean, = 180

b) Standard deviation of the sample mean, = = 15.8114

c) If the 10 people weigh more than 2400 pounds, Then each person have to weigh on average more than (2400/10) i.e, 240 pounds

P( Elevator is overweight) = P( > 240)

= P( > )

= P( z > 3.79)

= 1- P( z < 3.79)

= 1- 0.99992

= 0.00008

d) Sample size , n= 12

  Mean of the sample mean, = 180

Standard deviation of the sample mean, = = 14.4337

P( Elevator is overweight) = P( > 240)

= P( > )

= P( z > 4.16)

= 1- P( z < 4.16)

= 1- 0.99998

= 0.00002