Question

The amount of time part time worker at a pet store is normally distributed with a mean of 25 hours and a standard deviation of 2.5 hours. In a random sample of 25 employees, what is the probability that a worker will work for between 24 and 26 hours?

A student goes to the university library. The probability that the student checks out the science book for the science class is 0.40; the probability that the student checks out the math book is 0.30; and the probability that the student checks out both the science and math book is 0.20. What is the probability that the student checks out the science and the math book?

Dutch elm disease of American elm trees is caused by a fungus. It is known that 95% of elm trees in a city that have been treated with a fungicide recover from the disease. What is the probability that in a random sample of 4 elm trees in this city, at least 3 trees recover when treated with the fungicide?

A canning company claims that the amount of sugar in a cup of cherries is 13.2 grams. The population is normally distributed with a population standard deviation known to be 0.5 grams. The researcher took a sample of 25 cups of cherries and found a mean of 12.95 grams of sugar. Test at the 5% significance level that the amount of sugar in a cup of cherries is different than what the company claims. What is the p value for this test and what is your conclusion?

Answer 1

1.The amount of time part time worker at a pet store is normally distributed with a mean of 25 hours and a standard deviation of 2.5 hours. In a random sample of 25 employees, what is the probability that a worker will work for between 24 and 26 hours?

The amount of time part time worker has mean 25 and sd 2.5 hours. Also given that n=25.

We need to find out the Prob that a worker will work for between 24 and 26 hours:

We know that X~N(25,6.25)

To find out

We also know that follows a standard normal distribution.

Now when X=24,

  

When Z=26,

  

Therefore we need to find the

Which from the tables or from EXCEL we can find that 1-2*0.3446=0.3108.

The probability that a worker will work for between 24 and 26 hours=0.3108.

2. A student goes to the university library. The probability that the student checks out the science book for the science class is 0.40; the probability that the student checks out the math book is 0.30; and the probability that the student checks out both the science and math book is 0.20. What is the probability that the student checks out the science and the math book?

It is given that The probability the student checks out both the science and math book is 0.20, therefore the

the probability that the student checks out the science and the math book=0.20

3. Dutch elm disease of American elm trees is caused by a fungus. It is known that 95% of elm trees in a city that have been treated with a fungicide recover from the disease. What is the probability that in a random sample of 4 elm trees in this city, at least 3 trees recover when treated with the fungicide?

The probability of recovery of tree treated with fungicide 95% or 0.95.

Therefore at least 3 trees out of 4 tress will be 3 or 4 trees=Probability of 3 trees+probability of 4 trees.

  

  

  

A canning company claims that the amount of sugar in a cup of cherries is 13.2 grams. The population is normally distributed with a population standard deviation known to be 0.5 grams. The researcher took a sample of 25 cups of cherries and found a mean of 12.95 grams of sugar. Test at the 5% significance level that the amount of sugar in a cup of cherries is different than what the company claims. What is the p value for this test and what is your conclusion?

Test Statistic : follows .

The p-value is 0.0197 <0.5 and hence the null hypothesis is rejected. We conclude that the amount of sugar in a cup of cherries is different than what the company claims.