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
For each coin, there are <u>2 outcomes</u>, heads (H) or tails (T), so for the <u>3 coins</u>, the number of outcomes are:
2 x 2 x 2 = <u>8 outcomes:</u>
HHH / HHT / HTH / THH / TTT / TTH / THT / HTT
Out of all these outcomes, there are <u>3 ways</u>, we get exactly 2 Heads:
HHT / HTH / THH
So, the probability of exactly 2 coins landing on Heads is 3/8, or 0.375
If you multiply -222•10 what is the correct answer?
Answer:
- 12/7|x-9.6|+12.8
Step-by-step explanation: