# Np>5 and nq>5, estimate P(at least 11) with n=13 and p = 0.6 by using the normal distributions as an approximation to the binomial distribution. if np<5 or nq<5 then state the normal approximation is not suitable. p (at least 11) = ?

Question
Normal distributions
Np>5 and nq>5, estimate P(at least 11) with n=13 and p = 0.6 by using the normal distributions as an approximation to the binomial distribution. if $$\displaystyle{n}{p}{<}{5}{\quad\text{or}\quad}{n}{q}{<}{5}$$</span> then state the normal approximation is not suitable.
p (at least 11) = ?

2021-01-28
Step 1
Given n=12 and p=0.6.
Then,
Mean = np
$$\displaystyle={13}\times{0.6}$$
=7.8
Standard deviation $$\displaystyle=\sqrt{{{n}{p}{\left({1}-{p}\right)}}}$$
$$\displaystyle=\sqrt{{{7.8}{\left({1}-{0.6}\right)}}}$$
$$\displaystyle=\sqrt{{{3.12}}}$$
=1.7663
Step 2
So,
$$\displaystyle{P}{\left({X}\ge{11}\right)}={P}{\left(\frac{{{X}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\ge\frac{{{11}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\right)}$$
$$\displaystyle={P}{\left({Z}\ge\frac{{{11}-{7.8}}}{{{1.7663}}}\right)}$$
$$\displaystyle={P}{\left({Z}\ge{1.81}\right)}$$ From the right tailed z - table.
=0.4649
Therefore, p (at least 11)=0.4649

### Relevant Questions

A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
Basic Computation: hat p Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(b) Suppose $$n= 20$$ and $$p=0.23$$. Can we safely approximate the \hat{p} distribution by a normal distribution? Why or why not?
Basic Computation:$$\hat{p}$$ Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(b) Suppose $$n= 25$$ and $$p= 0.15$$. Can we safely approximate the $$\hat{p}$$ distribution by a normal distribution? Why or why not?
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(a) Suppose $$n = 100$$ and $$p= 0.23$$. Can we safely approximate the \hat{p} distribution by a normal distribution? Why?
Compute $$\mu_{hat{p}}$$ and $$\sigma_{hat{p}}$$.
Basic Computation:$$\hat{p}$$ Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(c) Suppose $$n = 48$$ and $$p= 0.15$$. Can we approximate the $$\hat{p}$$ distribution by a normal distribution? Why? What are the values of $$\mu_{hat{p}}$$ and $$\sigma_{p}$$.?
Basic Computation:$$\hat{p}$$ Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
(a) Suppose $$n = 33$$ and $$p = 0.21$$. Can we approximate the $$\hat{p}$$
distribution by a normal distribution? Why? What are the values of $$\mu_{hat{p}}$$ and $$\sigma_ {\hat{p}}$$.?
a) Which of the following properties distinguishes the standard normal distribution from other normal distributions?
-The mean is located at the center of the distribution.
-The total area under the curve is equal to 1.00.
-The curve is continuous.
-The mean is 0 and the standard deviation is 1.
b) Find the probability $$\displaystyle{P}{\left({z}{<}-{0.51}\right)}$$ using the standard normal distribution.
c) Find the probability $$\displaystyle{P}{\left({z}{>}-{0.59}\right)}$$ using the standard normal distribution.
For large value of n, and moderate values of the probabiliy of success p (roughly, $$0.05\ \Leftarrow\ p\ \Leftarrow\ 0.95$$), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation
$$\sigma = \sqrt{np(1\ -\ p)}$$. Explain this approximation making use the Central Limit Theorem.
A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
...