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:
P could be the price for working as a dog sitter where you ask for $40 to be payed every time you show to work so you can clean the hair off your clothes at the cleaner after work, plus $22 for every hour t you work .
Answer:
not enough
Step-by-step explanation:
only given 2 congruent sides and you need 3
Answer:
41.5675675676km/hour or 41.57km/hour
Step-by-step explanation:
158÷3.7=41.5675675676 or 41.57
720 ways... if you divide six into all answers, 720 is the only rational number
