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!

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

step-by-step explanation:

consider an orange peel: if you want to try and lay it flat, you have to stretch it, we know exactly how things are being stretched or squashed at any given point. we have many different map projections because each has different patterns of some projections can even preserve certain features of the earth without