Question

Question 1
“PhysicalFun” is a physical center which opens 24 hours per day. Each member of “PhysicalFun” has an access card. Members would be charged in a weekly basic (Sunday 12:00a.m. to Saturday 11: 59 p.m.) according to the accumulated number of hours of facilities used within a week. The weekly basic charge is $150 with an additional charge of $30 per hour. According to the company record, the number of hours a member spends in “PhysicalFun” in a week is normally distributed with mean 15 hours and standard deviation 2.8 hours.
(a) What are the average and standard deviation of weekly charge of a member?
(b) Suppose the middle 85% of weekly charge of a member is denoted by $(L1, L2). Find the values of L1
and L2.
(c) The senior management suggests fixing the weekly charge per member at $700. Assume the number
of hours a member spends in “PhysicalFun” would not be changed due to the change of the weekly charge
calculation method. What proportion of members would pay more money than the original system?
(d) How many hours does a member spend in “PhysicalFun” in a week so he / she would be beneficial by
the new system?

Answer 1

Answer:

Given that,

“PhysicalFun” is a physical center that opens 24 hours per day. Each member of “PhysicalFun” has an access card.

Members would be charged on a weekly basis (Sunday 12:00a.m. to Saturday 11: 59 p.m.) according to the accumulated number of hours of facilities used within a week.

The weekly basic charge is $150 with an additional charge of $30 per hour.

According to the company record, the number of hours a member spends in “PhysicalFun” in a week is normally distributed with mean 15 hours and a standard deviation of 2.8 hours.

(a).

What are the average and standard deviation of the weekly charge of a member:

Let the random variable, X = number of hours a member spends in "PhysicalFun" in a week.

The random variable X is normally distributed with,

Mean = 15 hours

Standard deviation = 2.8 hours.

Let the random variable, Y = weekly charges of a member spends in "PhysicalFun"

Where the random variable,

Y=150+ 30 X

The mean standard deviation values are obtained by using the properties of the random variable,

=150+30 15

=150+450

=600

=30 2.8

=84

(b).

Suppose the middle 85% of the weekly charge of a member is denoted by $(L1, L2). Find the values of L1 and L2:

The 85% limits for the mean is obtained as follows,

The number of standard deviation from the mean for the upper limit L2 is obtained using the standard normal distribution table (by calculating the Z- score),

P(Z > k)=0.075

k=1.4395

(in excel use function=1-NORM.S>INV(0.075))

The lower limit, L1==600-1.4395 84=479.0794

The upper limit, L2==600+1.4395 84=720.9206

(c).

What proportion of members would pay more money than the original system:

The senior management suggests fixing the weekly charge per member at $700.

Assume the number of hours a member spends in “PhysicalFun” would not be changed due to the change of the weekly charge calculation method.

The proportion value is obtained by calculating the Z-score,

P(Y > 700) P(Z >1.1905)

P(Y > 700) 1-P(Z 1.1905)

The probability value is obtained from the Z distribution tables for Z=21.1905 (In excel use function=1-NORM.S.DIST(1.1905, TRUE)).

P(Y > 700)=1-0.8831

=0.1170

The proportion of members=0.117

(d).

How many hours does a member spend in “PhysicalFun” in a week so he/she would be beneficial by the new system:

To get the benefit from the new system, the random variable Y should be less than $700.

Y=150+30 X < 700

X < (700-150)/30

X < 18.3333

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