Question

А B P c D E F G H K M N The average daily rate of a hotel in Canada as of August 2018 was $182.75. Assume the average daily rate follows a normal distribution with a standard deviation of $24.80 a. What is the probability that the average daily rate of a Canadian hotel will be: (i) less than $175 P(X < 175)= (ii) more than $200 POX > 200)= (iii) Between $150 and $195 P(150 < X < 195)= b. Determine the average daily rates that separate the (i) top 7% of average daily rates from the rest of the daily rates or from the bottom 93% of average daily rates x= Round to 2 decimal places (W) bottom 10% of average daily rates from the rest of the daily rates X Round 0 2 decimal places (iii) middle 65% of average daily rates from the rest of the daily rates <X Round to 2 decual places

Answer 1

Solution:-

Mean = 182.75, S.D = 24.80

a)

P(x < 175) = 0.3773

x = 175

By applying normal distribution:-

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z = - 0.3125

Use the z-score table or p-value calculator.

P(z < - 0.3125) = 0.3773

P(x > 200) = 0.7567

x = 200

By applying normal distribution:-

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z = 0.6956

Use the z-score table or p-value calculator.

P(z > 0.6956) = 0.7567

b)

i) The top 7% of average daily rates from the rest of the daily rates is 219.35.

p-value for the top 7% = 0.93

z-score for the p-value = 1.476

By applying normal distribution:-

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x = 219.35

ii) The bottom 10% of average daily rates from the rest of the daily rates is 150.96

p-value for the bottom 10% = 0.10

z-score for the p-value = -1.282

By applying normal distribution:-

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x = 150.96

iii) The middle 65% of average daily rates from the rest of the daily rates is 159.56 and 205.94.

p-value for the middle 65% = 0.175 and 0.825

z-score for the p-value = - 0.935 and + 0.935

By applying normal distribution:-

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x₁ = 159.56

x₂ = 205.94