Answer:
a.0.8664
b. 0.23753
c. 0.15866
Step-by-step explanation:
The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between
a. N164.50 and N185.50?
b. greater than N180.00?
c. less than N168.00?
We solve the above question using z score formula
z = (x-μ)/σ/√n where
x is the raw score,
μ is the population mean = N175
σ is the population standard deviation = N42
n is random number of sample = 36
a. Between N164.50 and N185.50?
For x = N 164.50
z = 164.50 - 175/42 /√36
z = -1.5
Probability value from Z-Table:
P(x = 164.50) = 0.066807
For x = N185.50
z = 185.50 - 175/42 /√36
z =1.5
Probability value from Z-Table:
P(x=185.50) = 0.93319
Hence:
P(x = 185.50) - P(x =164.50)
= 0.93319 - 0.066807
= 0.866383
Approximately = 0.8664
b. greater than N180.00?
x > N 180
Hence:
z = 180 - 175/42 /√36
z = 5/42/6
z = 5/7
= 0.71429
Probability value from Z-Table:
P(x<180) = 0.76247
P(x>180) = 1 - P(x<180) = 0.23753
c. less than N168.00?
x < N168.
z = 168 - 175/42 /√36
z = -7/42/6
z = -7/7
z = -1
Probability value from Z-Table:
P(x<168) = 0.15866
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ±
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71

[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
Tim spends 500 minutes at the school.
Step-by-step explanation:
Given,
Time for lunch = 45 minutes
Time for study hall = 45 minutes
Total = 45+45 = 90 minutes
This represents 18% of the total minutes Tim spend at school.
Let,
x be the total minutes Tim spend at school.
18% of x = 90 minutes

Dividing both sides by 0.18

Tim spends 500 minutes at the school.
Keywords: percentage, division
Learn more about division at:
#LearnwithBrainly
Answer:
1749
Step-by-step explanation:
You would set up a proportion that says 901/x = 34/100. You would cross multiply to get that 34x = 90100. You would divide 90100 by 34 to get 2650. Then do 2650 - 901 to get 1749.