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:
Dan bought 8 shirts and 3 pants. This means that the ratio representing the number of shirts to the number of pants that Dan bought is 8/3 = 2.67
Devonte bought 12 shirts and 5 pants. This means that the ratio representing the number of shirts to the number of pants that Devonte bought is 12/5 = 2.4
Therefore, they are not equivalent ratios because the relationship is not exactly the same.
First off, let's find the 1st term's value, and the 30th term's value,
