Answer:
Z= 0.253
Z∝/2 = ± 1.96
Step-by-step explanation:
Formulate the null and alternative hypotheses as
H0 : u1= u2 against Ha : u1≠ u2 This is a two sided test
Here ∝= 0.005
For alpha by 2 for a two tailed test Z∝/2 = ± 1.96
Standard deviation = s= 15
n= 10
The test statistic used here is
Z = x- x`/ s/√n
Z= 2058- 2046 / 15 / √10
Z= 0.253
Since the calculated value of Z= 0.253 falls in the critical region we reject the null hypothesis.
There is evidence at the 0.05 level that the doors are too short and unusable.
Answer:
80
Step-by-step explanation:
x³ + x²
4³ + 4²
64 + 16
80
Answer:
The probability that the average score of the 49 golfers exceeded 62 is 0.3897
Step-by-step explanation:
The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5


We are supposed to find he probability that the average score of the 49 golfers exceeded 62.
Formula : 


Refer the z table for p value
p value = 0.6103
P(x>62)=1-P(x<62)=1-0.6103=0.3897
Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897
Answer:
5pi
Step-by-step explanation: