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
Answer:
........................
Step-by-step explanation:
This can be solved using the formula for t test which is = x - mu over standard deviation divided by the square root of the number of samples. In this case, it is 1185 - 1200 over 70 divided by square root of 100 which give us a value of 2.14 which is equal to .9839 in area. Only rarely, just over one time in a hundred tries of 100 light bulbs, would the average life exceed 1200 hours.
Y= Mx+b
In the equation of a straight line,the slope is the number "m" that is multiplied on the x, and "b<span>" is the </span>y<span>-intercept (that is, the point where the line crosses the vertical </span>y<span>-axis).</span>
