286 ÷ 5 1/2 hours = 286 ÷ 5.5 = 52 final answer
Answer:
192y^8
Step-by-step explanation:
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:
The means differ by 1, but the ranges differ by 40.
Step-by-step explanation:
The mean for LaTesha's score is (92+45+67+36+80)/5= 64
The mean for Benards score is (63+68+62+69+53)/5= 63
The range for LaTeshas score is 92-36=56
The range for Benards score is 69-53=16
So, 64-63=1 and 56-16= 40
Answer:
Step-by-step explanation:
