Answer:
idk
Step-by-step explanation:
Answer:
the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Step-by-step explanation:
since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:
Z= S²*(n-1)/σ²
follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom
since
S² > 3.10 , σ²= 1.75 , n= 20
thus
Z > 33.65
then from χ² distribution tables:
P(Z > 33.65) = 0.02020
therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Answer:
Step-by-step explanation:
Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as
z = (x - u)/s
Where
x = ticket sales from monday
u = mean amount of ticket
s = standard deviation
From the information given,
u = 500 tickets
s = 50 tickets
We want to find the probability that the mean will be greater than 510. It is expressed as
P(x greater than 510) = 1 - P(x lesser than or equal to 510)
For x = 510
z = (510 - 500)/50 = 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.9773
P(x greater than 510) = 1 - 0.9773 = 0.0227
Perimeter P = 2L + 2W
b. P = 2*1010 + 2*1818 = 2020 + 3636 = 5656 inches
c. Area A = WL
d Area = 1010*1818 = 1836180 in^2
Answer:
55.9%
Step-by-step explanation:
To find the percent that are girls, take the number of girls over the total number of students
19/34
.558823529
To change to percent form, multiply by 100%
55.8823529%
To 1 decimal place
55.9%
