The daily dinner bib in a local restaurant are normally dist

The daily dinner bib in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. Define the random variable in words. What is the probability that a randomly selected bill will be at least $39.10? What percentage of the bills will be less than $16.90? What are the minimum and maximum of the middle 95% of the bills? If twelve of one day\'s bilk had a value of at least S43.06, how many bills did the restaurant collect on that day?

Solution

Part A)

Random variable:

A random variable is defined as any variable or any value which is subject to change that means which keep on changing and the change cannot be predicted in advance with 100% accuracy.

for example: temperature is an example of random variable because it keeps on changing and the next coming value of the temperature cannot be predicted in advance with 100% accuracy. We can predict the value of temperature but that will not be 100% accurate.

consider an another example: The blood pressure is an example of random variable because we cannot predict in advance, that what will be next value of blood pressure and it keeps on changing after every second, minutes or hour.

Some other examples of blood pressure

Speed of a vehicle, Stock prices, Humidity, air pressure.

----------------------------------------------------------------------------------------------------------

Part b)

Let the name of variable be X

X represents a continuous random variable representing the amount of bill.

X ~ N(28, 6)        

This notation above says that X approaches a normal distribution (population) with population mean 28 and standard deviation of 6

We have to find

P(X 39.10)

Z Score = (x - µ)/

µ: Population mean

: Population standard deviation

P(Z (39.10 - 28)/6)

P(Z 1.85)

Now we will refer to the standard normal table to find the probability.

P(Z 1.85) = 1 – P(Z 1.85)

= (1 – 0.96784) = 0.03216

So the Probability is 0.03216 or 3.216 % that bill will be at least $39.10

--------------------------------------------------------------------------------------------------------------------------------

Part C)

What Percentage of bill will be less than $16.9?

P(X < 16.9)

Z Score = (x - µ)/

µ: Population mean

: Population standard deviation

P(Z < (16.9 - 28)/6)

P(Z < - 1.85)

Now we will refer to the standard normal table to find the probability.

P(Z - 1.85)= 1 – P(Z < 1.85)

= (1 – 0.96784) = 0.03216

So the Probability is 0.03216 or 3.216 % that bill will be less that $16.9

--------------------------------------------------------------------------------------------------------------

Part d)

What are the minimum and maximum of middle 95% of bill?

As per the standard normal distribution table the middle 95% data is covered between

-1.96 to +1.96

P(-1.96 < Z < 1.96) = 0.95

Z Score = (x - µ)/

Minimum value of bill will be as follows

-1.96 = (X – 28)/6

(X – 28) = -11.76

X = 28 – 11.76

X = 16.24

Maximum value of Bill will be

1.96 = (X – 28)/6

(X – 28) = 11.76

X = 28 + 11.76

X = 39.76

So the value from ($ 16.24 to $ 39.76) will cover the middle 95% of data.

-------------------------------------------------------------------------------------------------------------------------------------

Part e)

If 12 of one day bill had value of at least $ 43.06, how many bills did the restaurant collect on that day

First we find the Probability of any bill greater than or equal to 43.06

P(Z (43.06 - 28)/6)

P(Z 2.51)

Now we will refer to the standard normal table to find the probability.

P(Z 2.51) = 1 – P(Z 2.51)

= (1 – 0.99396) = 0.00604

So the Probability is 0.00604 that bill will be at least $43.06

So total number of bills collected on that day will be

12/0.00604 = 1986.75 which is approximately 1987 bills

Because if 12 bills have bill greater than or equal to 43.06 that means remaining bill will be lower than 43.06

For getting 0.00604 probability we need 12 bills

For getting full 1 probability we need (12/0.00604) = 1986.75 or 1987 bills

End of the answer

Thank you