Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.
<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 binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with
.
In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:


The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:


Z = -2.3
Z = -2.3 has a p-value of 0.0107.
0.0107 = 1.07% probability that more than 30 are single.
More can be learned about the normal distribution at brainly.com/question/24663213
Answer:
paul bc he could not get george on the phone
Explanation:
Answer:
Hola!
Explanation:
¡Gracias por los puntos! ¡Realmente lo aprecio! ¡Que tengas un gran día, eres increíble!
Realmente no puedo hablar español * pequeña risa
:)
The Confidence Interval of the given random sample survey is gotten as; CI = 13.755 ± 1.456
<h3>What is the confidence Interval?</h3>
We are given the hypotheses as;
Null Hypothesis; H₀: μ = 15
Alternative Hypothesis; Hₐ: μ ≠ 15.
where μ is the mean number of hours worked per week for all students at the college with part-time jobs.
We are told that it has a 95% confidence interval and degree of freedom of df = 25
From the t-table, the t-score = 2.06
The table gives us the standard error as 0.707 and sample mean of 13.755. Thus, we can find the confidence interval as;
CI = 13.755 ± 2.06(0.707)
CI = 13.755 ± 1.456
Read more about Confidence Interval at; brainly.com/question/17097944