Given:
population mean, μ =135
population standard deviation, σ = 15
sample size, n = 19
Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.
Estimate of sample mean
Estimate of sample standard deviation
to 5 decimal places.
Thus, using the normal probability table,
Therefore
The probability that the mean weight is between 125 and 130 lbs
P(125<X<130)=0.0731166-0.0018308
=
0.0712858
Answer:
-472x+ 177
Step-by-step explanation:
-(8x-3)59
-59(8x-3)
-59*8x-(-59)*3
-59*8x+59*3
Answer: y=3x+12
explanation: just add the 3x to the 12
Answer: (23.53, 25.47)
Step-by-step explanation:
The confidence interval for population mean(when population standard deviation is unknown) is given by :-
, where 
= sample mean
n = sample size.
s = sample standard deviation.
t* = Critical value.
Given : 
s= 4.4
Confidence level = 90% =0.090
Significance level = 
Sample size : n= 58
Degree of freedom : df= n-1= 57
Using t-distribution table , the critical value :
Then, the confidence interval will be :-
Hence, a 90% confidence interval for the population mean = (23.53, 25.47)
1/2 inch
Explanation
I had this question on my homework and got it right. Hope this helps!
