Question

A cab driver knows from experience that the number of fares he will pick up in an evening is a random variable with u = 21 and standard deviation = 3.4. Assuming that the distribution of this random variable can be approximated closely with a normal curve, find the probabilities that in an evening the driver will pick up (a) at least 30 fares (b) anywhere from 20 to 25 fares (including 20 and 25). Since the number of fares is a discrete random variable, and we are using a normal curve, we will need continuity corrections on both parts a and b above. Those continuity corrections are given below. (a) P (at least 30 fares) = P (X greater than or equal to 30) = P (X>29.5) after continuity correction. Please find this probability. (b) P(between 20 and 25 fares inclusive) = P(19.5

Answer 1

Answer:

a. P(X ≥ 30) = 0.004

b. P(20 ≤ X ≤ 25) = 0.49509

a'. P(X > 29.5) = 0.00621

b'. P(19.5 ≤ X ≤ 24.5) = 0.51852

Step-by-step explanation:

Hello!

Your study variable is X: the number of fares the taxi driver will pick up in an evening.

This variable has an approximately normal distribution X≈N(μ;δ²)

Where

the population mean μ= 21

population standard deviation δ= 3.4

The probabilities asked are for a number that the variable can take X=x₀, so to calculate them you have to use the following standardization:

Z = (X - μ)/δ ≈ N(0;1)

a. (...)at least 30 fares.

Symbolically:

P(X ≥ 30)

Now you standardize it and calculate the value of Z to look for the probability in the Z-table

P(Z ≥ (30-21)/3.4) = P(Z ≤ 2.647) ≅ P(Z ≤ 2.65)

Since the Z-table has the values of the cumulative probabilities P(Z < Z$_{\alpha }$)= α you need to look for the cumulative probability up to the desired number and subtract it to the max probability (1).

P(Z ≤ 2.65) = 1 - P(Z > 2.65) = 1 - 0.9960 = 0.004

b.(...)anywhere from 20 to 25 fares.

This means that you want to know the probability of X being between 20 and 25.

Symbolically:

P(20 ≤ X ≤ 25)

You can rewrite it as:

P(X ≤ 25 ) - P(X ≤ 20)

Now you can standarize them and look for the cummulative probabilities in the table:

P(Z ≤ (25 - 21)/3.4 ) - P(Z ≤ (20-21)/3.4) = P(Z ≤ 1.18 ) - P(Z ≤ (-0.29)

= 0.88100 - 0.38591 = 0.49509

With the continuity corrections:

a'. P(X > 29.5) = P(Z > (29.5 - 21)3.4) = P(Z > 2.5)

= 1 - P(Z ≤ 2.5) = 1 - 0.99379 = 0.00621

b'. P(19.5 ≤ X ≤ 24.5) = P(X ≤ 24.5) - P(X ≤ 19.5)

= P(Z ≤ (24.5 - 21)/3.4) - P(Z ≤ (19.5-21)/3.4)

= P(Z ≤ (1.03) - P(Z ≤ (-0.44) = 0.84849 - 0.32997 = 0.51852

I hope you have a SUPER day!