Using the <em>normal distribution and the central limit theorem</em>, we have that:
a) A normal model with mean 0.3 and standard deviation of 0.0458 should be used.
b) There is a 0.2327 = 23.27% probability that more than one third of this sample wear contacts.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean
and standard deviation
, the z-score of a measure X is given by:

- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, for a <u>proportion p in a sample of size n</u>, the sampling distribution of sample proportion is approximately normal with mean
and standard deviation
, as long as
and
.
In this problem:
- 30% of students at a university wear contact lenses, hence p = 0.3.
- We randomly pick 100 students, hence n = 100.
Item a:


Hence a normal model is appropriated.
The mean and the standard deviation are given as follows:


Item b:
The probability is <u>1 subtracted by the p-value of Z when X = 1/3 = 0.3333</u>, hence:

By the Central Limit Theorem



has a p-value of 0.7673.
1 - 0.7673 = 0.2327.
0.2327 = 23.27% probability that more than one third of this sample wear contacts.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213
Well if you Multiply 5.4 times 5 yo get 27, so your answer should be 27in
Answer:
The minimum score required for an interview is 77.252
Step-by-step explanation:
We solve this using z score formula
z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
Top 15% of the candidates is a ranking that is equivalent to = 100 - 15% = 85th percentile.
The z score of 85th percentile = 1.036
Mean = 70
Standard Deviation = 7
Minimum score = raw score = ???
Hence:
1.036 = x - 70/7
Cross Multiply
1.036 × 7 = x - 70
7.252 = x - 70
x = 70 + 7.252
x = 77.252
The minimum score required for an interview is 77.252
Researcher
Military and airline forces
Teaching
Storm chaser
Environmental scientist