Question

Please help me if you can calculate it

I can only calculate Q1 a b c

Question 1
A taxi company reveals that the daily earnings of taxi drivers in the company follows a normal distribution
with a mean of $1062.5 and standard deviation of $350.
(a) Find the probability that a taxi driver earns less than $1500 in a day.
(b) 93.7% drivers earn more than $k a day. Find the value of k.
John is a taxi driver in this company. Besides driving taxi, he has another part time online job. The daily
earnings from the part time online job follow a normal distribution with a mean of $235.5 and standard
deviation of $84.5. The daily earnings from driving taxi and part time online job are assumed to be
independent.
(c) Use T to denote the total daily earning for a day he drives taxi and works on the part time online job.
Find the mean and standard deviation of T. (Round off the mean and standard deviation to the nearest
integer.)
(d) Hence, by using $(L1, L2) to denote the middle 96.6% of the total daily earning T, find the values of L1
and L2. (Round off L1 and L2 to the nearest integer.)
Question 2
Tom is a supermarket manager. He reviewed transaction time when a customer paid by credit card. The
transaction time is normally distribution with mean of 20 seconds and standard deviation of 5 seconds.
(a) For a group of 6 customers, find the probability that 5 customers can finish the transaction within 20
seconds. (Assume that the transaction times of customers are independent.)
After discussion with the network provider, he will upgrade the network so that it is promised that each
transaction time can be reduced by 15%.
(b) Use Y to denote the transaction time after network upgrade. Find the mean and standard deviation of Y.
(c) Calculate the 97th percentile of Y. (i.e. find the value of t such that P(Y (d) Compare with the transaction time before upgrade, is it (I) a higher proportion, (II) a lower proportion,
or (III) the same proportion of all customers can finish the transaction within 20 seconds? (Just state
your answer, no calculation is needed.)

Answer 1

Question 1:

(a) To find the probability that a taxi driver earns less than $1500 in a day, we need to standardize the value using the given mean and standard deviation, and then find the corresponding probability from the standard normal distribution table:

z = (1500 - 1062.5) / 350 = 1.20

Using the standard normal distribution table, the probability of a standard normal random variable being less than 1.20 is approximately 0.8849. Therefore, the probability that a taxi driver earns less than $1500 in a day is approximately:

P(X < 1500) = P(Z < 1.20) = 0.8849

(b) We need to find the value of k such that 93.7% of the drivers earn more than $k a day. This means that the probability of a driver earning less than or equal to $k a day is 1 - 0.937 = 0.063. We can standardize k using the given mean and standard deviation, and then find the corresponding z-score from the standard normal distribution table:

z = (k - 1062.5) / 350

Using the standard normal distribution table, we find that the z-score corresponding to a probability of 0.063 is approximately -1.51. Therefore:

-1.51 = (k - 1062.5) / 350

k = -1.51 * 350 + 1062.5 = $499.25 (rounded to the nearest cent)

(c) The mean of the total daily earning is:

μT = μ1 + μ2 = 1062.5 + 235.5 = 1298

The variance of the total daily earning is the sum of the variances of the two earnings, since they are assumed to be independent:

σT² = σ1² + σ2² = 350² + 84.5² ≈ 128681

Therefore, the standard deviation of the total daily earning is:

σT ≈ √128681 ≈ 358.5

(rounded to the nearest integer)

(d) To find L1 and L2, we need to find the z-scores corresponding to the lower and upper 2.2% tails of the standard normal distribution:

z1 = -1.81

z2 = 1.81

Then we can use the formula for standardizing a normal random variable to find the corresponding values of T:

z1 = (L1 - μT) / σT

z2 = (L2 - μT) / σT

Solving for L1 and L2, we get:

L1 = μT + z1σT ≈ 1298 + (-1.81) * 358.5 ≈ $645

L2 = μT + z2σT ≈ 1298 + 1.81 * 358.5 ≈ $1951

(rounded to the nearest integer)

Question 2:

(a) We can model the transaction time of a single customer as a normal random variable with mean 20 and standard deviation 5. Then the total transaction time for 6 customers can be modeled as a normal random variable with mean 6 * 20 = 120 and standard deviation √(6 * 5²) = 15. To find the probability that 5 customers can finish the transaction within 20 seconds, we need to standardize the value using this mean and standard deviation, and then find the corresponding probability from the standard normal distribution table:

z = (5 * 20 - 120) / 15 = -0.53