The probabilities and the number of automobiles lined up at a Lakeside Olds at opening time (7:30 a. m.) for service arecol1 Number 1 2 3 4col2 Probability 0.40 0.03 0.06 0.51On a typical day, how many automobiles should Lakeside Olds expect to be lined up at opening?

EX=2.68 ≈ 3

Lakeside Olds should expect 3 automobiles lined up at opening

Step-by-step explanation:

Expected Value of a Discrete Probability Distribution

Given a discrete probability distribution of values

x={x1,x2,x3...,xn}

And probabilities

p={p1,p2,p3,...,pn}

Provided the sum of all probabilities is 1, then the expected value of the distribution is

EX =

The data refers to the number of automobiles lined up at Lakeside Olds at opening time for service:

x={1,2,3,4}

And probabilities

p={ 0.40 , 0.03 , 0.06 , 0.51 }

Checking the sum: 0.40 + 0.03 + 0.06 + 0.51 = 1

Now compute the expected value

EX= 1*0.40+2*0.03+3*0.06+4*0.51

EX=2.68 ≈ 3

Lakeside Olds should expect 3 automobiles lined up at opening

question:

could we have options?

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.