Question

1- Assume that visitors of a hotel on average pay $20 for minibar per night per room, with a standard deviation of $3. Assume further that minibar expenses are normally distributed.
a- What percentage of rooms are expected to pay more than $25 per night, i.e. P(x > 25)
b- What percentage of rooms are expected to pay more than $40 per night, i.e. P( x > 40)?
c- What percentage of rooms are expected to pay less than $12 per night, i.e. P( x < 12)?
d- What percentage of rooms are expected to pay between $18 and $24, i.e. P(18 < x < 24)?
e- What percentage of rooms are expected to pay between $16 and $19, i.e. P (16 < x < 19)?

Answer 1

This is a normal distribution question with

a) x = 25

P(x > 25.0)=?

The z-score at x = 25.0 is,

z = 1.6667

This implies that

P(x > 25.0) = P(z > 1.6667) = 1 - 0.9522129635397043

P(x > 25.0) = {0.0478}=4.78%

b) x = 40

P(x > 40.0)=?

The z-score at x = 40.0 is,

z = {40.0-20.0}/{3.0}

z = 6.6667

This implies that

P(x > 40.0) = P(z > 6.6667) = 1 - 0.999999999986919

P(x > 40.0) = {0.0}= 0 %

c) x = 12

P(x < 12.0)=?

The z-score at x = 12.0 is,

z = \frac{12.0-20.0}/{3.0}

z = -2.6667

This implies that

P(x < 12.0) = P(z < -2.6667) = \textbf{0.003830000718271452}=0.38%

d) x1 = 18

x2 = 24

P(18.0 < x < 24.0)=?

This implies that

P(18.0 < x < 24.0) = P(-0.6667 < z < 1.3333) = P(Z < 1.3333) - P(Z < -0.6667)

P(18.0 < x < 24.0) = 0.9087833131501531 - 0.25248188939838734

P(18.0 < x < 24.0) = \textbf{0.6563}=65.63%

e) x1 = 16

x2 = 19

P(16.0 < x < 19.0)=?

This implies that

P(16.0 < x < 19.0) = P(-1.3333 < z < -0.3333) = P(Z < -0.3333) - P(Z < -1.3333)

P(16.0 < x < 19.0) = 0.36945391969257046 - 0.09121668684984685

P(16.0 < x < 19.0) = \textbf{0.2782}=27.82%

PS: you have to refer z score table to find the final probabilities.

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