Note :- If sample size n > 30 OR Population standard deviation σ is given then We use Z test.
If sample size n < 30 AND Population standard deviation σ is unknown then we use t test.
In this question we are not given Population standard deviations σ1 and σ2.
Also n1 = 14 and n2 = 13 both sample sizes are LESS than 30.
Therefore we use t test.
Given :- Assume the population standard deviations are equal. ( σ1 = σ2)
then Degrees of freedom = n1 + n2 - 2 = 14 + 13 - 2 = 25.
(kindly find the image attached with this solution)
Answer :- Therefore 25 degrees of freedom are used to find the t critical value.
6(2x-11)+15=3x+12 Given
12x-66+15=3x+12 Distribution
12x-51=3x+12 Combine like terms
12x=3x+63 addition
9x=63 subtraction
x=7 division
the value of x that makes the equation true is 7.
The answer is 8shares
Step- My - Step
2,000 divided by 237.68
= 8.414....
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
