Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
190. Count it out on your fingers, starting with -10,-15, and -20. Then count out both hands.
Let's assign variables
The time required to <span>wire a building: y
</span>
The number of electricians <span>to have done this job: x
</span>
<span>y varies inversely as x
so that:

where k is the proportional constant
To calculate the value of k, plug in the values of x and y:
x=7
y=4

Now, write down the relation between x and y:

"</span><span>how long would it have taken 3 electricians to have done the job?"
x=3
Let's find out y:

So it would take 9 hours and 20 minutes to have this job done.
</span><span />