The answer to your question is w =40(2)n [n is in exponent form I just couldn’t type it that way ]
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:
TP = 5. greater than
Step-by-step explanation:
1. Out of the 4 options, HH, HT, TH, TT, there is one option we want (HT) which is one our of the four options. that means that the theoretical probability is 1/4 = 25%. Since there were 20 flips, the theoretical probability is 25% of 20 which is 5.
2. For experimental probability, its what actually happenned. Out of the 20 flips, 6 were HT so comapared to the Theoretical probability, the experimental probability was higher.
15.6 dollars s how much money Wendell spent
