I need help I'm lost !!
1 answer:
You might be interested in
Answer:
And we can find this probability with this difference:
And in order to find these probabilities using the normal standard distribution or excel and we got.
So we expect about 99.735% of values between $29 and $89
Step-by-step explanation:
Let X the random variable that represent the amount of Jens monthly phone of a population, and for this case we know the distribution for X is given by:
Where and
We are interested on this probability first in order to find a %
The z score is given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
And in order to find these probabilities using the normal standard distribution or excel and we got.
So we expect about 99.735% of values between $29 and $89
By assuming the standard deviation of population 2.2 the confidence interval is 8.67 toys,8.94 toys.
Given sample size of 1492 children,99% confidence interval , sample mean of 8.8, population standard deviation=2.2.
This type of problems can be solved through z test and in z test we have to first find the z score and then p value from normal distribution table.
First we have to find the value of α which can be calculated as under:
α=(1-0.99)/2=0.005
p=1-0.005=0.995
corresponding z value will be 2.575 for p=0.995 .
Margin of error=z*x/d
where x is mean and d is standard deviation.
M=2.575*2.2/
=0.14
So the lower value will be x-M
=8.8-0.14
=8.66
=8.67 ( after rounding)
The upper value will be x+M
=8.8+0.14
=8.94
Hence the confidence interval will be 8.67 toys and 8.94 toys.
Learn more about z test at brainly.com/question/14453510
#SPJ4