A 99% confidence interval for the population mean of high school students that take the bus to school every day is b) ci=(44.34%, 54.43%)
We need to find the 99% confidence interval for the population mean of high school students that take the bus to school everyday
A confidence interval is a range of estimates for an unknown parameter. The confidence interval is calculated at the specified confidence level; the most common is the 95% confidence level, but sometimes other levels are used, such as 90% or 99%.
The confidence interval of proportions is given by:
π ± z √(π (1-π) /n)
π is the sample proportion.
z is the critical value.
n is the sample size.
For 99% confidence interval the value of z is 2.58
π = 321/650
The confidence interval is given by
=321/650 ± 2.58 √( (321/650) × [ 1 - (321/650) ] ÷ 650)
= (0.493846 ± 0.050594)
=(0.4434 , 0.5443)
=(44.34 % , 54.43 %)
Hence a 99% confidence interval for the population mean of high school students that take the bus to school every day is ci=(44.34%, 54.43%)
<u>Learn more about confidence interval:</u>
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Answer:
Step-by-step explanation:
m = 4 ; (1,-2)
y - y₁ = m (x - x₁)
y - [-2] = 4 (x - 1)
y + 2 = 4x - 4
y = 4x - 4 - 2
y = 4x - 6
Answer:
the answer is a i got it right
Step-by-step explanation:
Using the normal distribution, we have that:
The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean 
and standard deviation 
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation of men's heights are given as follows:
.
The proportion of men who meet the height requirement is is the <u>p-value of Z when X = 62 subtracted by the p-value of Z when X = 55</u>, hence:
X = 62:
Z = -1.87
Z = -1.87 has a p-value of 0.0307.
X = 55:
Z = -3.67
Z = -3.67 has a p-value of 0.0001.
0.0307 - 0.0001 = 0.0306 = 3.06%.
The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.
More can be learned about the normal distribution at brainly.com/question/4079902
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