A bank has kept records of the checking balances of its cust
A bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $300 with a standard deviation of $48. A random sample of 144 checking accounts is selected.
1.
What is the probability that the sample mean will be more than $306.60?
2.
What is the probability that the sample mean will be less than $308?
3.
What is the probability that the sample mean will be between $302 and $308?
4.
What is the probability that the sample mean will be at least $296?
| 1. | What is the probability that the sample mean will be more than $306.60? |
| 2. | What is the probability that the sample mean will be less than $308? |
| 3. | What is the probability that the sample mean will be between $302 and $308? |
| 4. | What is the probability that the sample mean will be at least $296? |
Solution
let Xbar be the sample mean of a random sample of size 144 from the random variable X
where X denotes the daily balances of its customers
given that X~N($300,482)
so Xbar~N(300,482/144=16=42)
so mean of Xbar is 300 and variance is 16 and SD is 4
1. P[Xbar>306.6]=1-P[Xbar<306.6]=1-P[(Xbar-300)/4<(306.6-300)/4]=1-P[Z<1.65]=1-0.950529=0.049471 [answer]
2. P[Xbar<308]=P[(Xbar-300)/4<(308-300)/4]=P[Z<2]=0.977250 [answer]
3.P[302<Xbar<308]=P[(302-300)/4<(Xbar-300)/4<(308-300)/4]=P[0.5<Z<2]=P[Z<2]-P[Z<0.5]=0.977250- 0.691462=0.285788 [answer]
4. P[Xbar>296]=1-P[(Xbar-300)/4<(296-300)/4]=1-P[Z<-1]=1-0.158655=0.841345 [answer]
[Z~N(0,1)]
