Question

A hotel in Saint Augustine books an average of 50 rooms per night. There is a standard deviation of 10 rooms per night. What is the probability that...

a) more than 50 rooms are booked on a given night?

b) between 50 and 60 rooms are booked on a given night?   

c) 60 rooms or less are booked on a given night?   

d) between 30 and 70 rooms are booked on a given night?

e) less than 40 rooms are booked on a given night?

Answer 1

Given:

= 50 rooms per night, = 10 rooms per night

Let X be the number of rooms booked per night

X follows Normal distribution.

Then, According to the Central Limit Theorem

a)

Find: P(X > 50)

P(X > 50) = P(Z > 0)

P(X > 50) = 1 - P(Z < 0)

P(X > 50) = 1 - 0.5 .....................Using standard Normal table

P(X > 50) = 0.5

b)

Find: P(50 < X < 60)

P(50 < X < 60) = P(0 < Z < 1)

P(50 < X < 60) = P(Z < 1) - P(Z < 0)

P(50 < X < 60) = 0.8413 - 0.5   .....................Using standard Normal table

P(50 < X < 60) = 0.3413

C)

Find: P(X < 60)

P(X < 60) = P(Z < 1)

P(X < 60) = 0.8413 .....................Using standard Normal table

D)

Find: P(30 < X < 70)

P(30 < X < 70) = P(-2 < Z < 2)

P(30 < X < 70) = P(Z < 2) - P(Z < -2)

P(30 < X < 70) = 0.9772 - 0.0228   .....................Using standard Normal table

P(30 < X < 70) = 0.9545

e)

Find: P(X < 40)

P(X < 40) = P(Z < -1)

P(X < 40) = 0.1587 .............................Using standard Normal Table