Use the same information from the problem 1: The average waiting time for a drive-in window at a local bank is 8.3 min. with SD of 2.4 min. Assuming that data is normally distributed, find the probability that the customer will have to wait less than 5 min. or more than 7 min. Note: make a picture of a bell-shaped distribution, show all information on it.

Step-by-step explanation:

Hello!

You have the information of a variable X: waiting time for a drive-in window at a local bank.

This variable has a normal distribution X~N(μ;σ²) with mean μ= 8.3min and standard deviation σ= 2.4min.

You need to calculate the probability of the customer waiting less than 5 minutes, symbolically: P(X<5) or more than 7 min, symbolically: P(X>7)

See distribution graph in attachment.

To calculate these probabilities you have to use the standard normal distribution: Z= (X - μ)/σ ~N(0;1) and get the values from the Z-tables.

P(X<5)

P(Z<(5-8.3)/2.4)

P(Z<-1.375)= 0.085

The tables give values of cummulative probabilities P(Z≤Zₐ), so to calculate values greater than a given Zₐ, you have to do the following:

P(X>7)

P(Z>(7-8.3)/2.4)

P(Z>-0.542)

1 - P(Z≤-0.542)= 1 - 0.294= 0.706

I hope it helps!

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

12

step-by-step explanation: