Given that the mean is $9.5 and the standard deviation is $1.30, the standard error will be given by:
σ/√n
where
σ-standard deviation
n=sample size
thus, we shal have:
1.30/√20
=0.2906
Next we find the margin error
0.2906*2=0.581
thus the confidence interval will be:
(9.5+0.581, 9.5-0.581)
=(10.081,8.919)
Answer:
The probability that a person is younger than 50 given that she uses onling banking is 1.5%
Step-by-step explanation:
The survey indicated that 50% of adults conducts their banking online.
P(online banking customers) =1/2
Of the total number of customer that bank online, the survey stated that 20% of them are younger than 50 years.
P(customers younger than 50 yrs) =1/5
Out of the 20% than are younger than 50, we were told that 15% of them conduct online banking.
P(younger than 50 online banking customer) =3/20
Therefore the the probability that a = person is younger than 50 given that she uses online banking is P = 3/20*1/5*1/2 =17==3/200 = 0.015 =1.5%
The probability that she is younger than 50 does not increase or decrease with additional information, it only makes it clearer.
25%
Hers how 100 divided by the denominator multiplied by the numerator will give u an answer in the percentage of any fraction so 100 divided by 4 = 25 multiplied by one = 25
(3y + 7) + (5y-7)= 8y
(4y - 5)- 8y= -4y-5
