Emerson’s math test scores are given in the table below:
87, 93, 92, 25, 96
a) Find the median.
b) Find the sample mean.
c) Find the sample standard deviation.
Question 2: Answer the following questions.
a) An animal shelter has a 58% adoption rate for puppies.
Of all puppies in the shelter, 80% live to be 7 years or older.
Of the puppies who are adopted, 90% live to be 7 years or
older. What is the probability that a randomly selected puppy in the
shelter will get adopted and live 7 or more years?
The sample space is the set of all pupies living in the shelter.
Let A be the event that a puppy is adopted, and O the event that a puppy lives to be 7 years or older. We are given, P(A)=.58, P(O)=.8, and P(O∣A)=.9, and we are asked to find P(A and O). We will use the formula
P(A and O)=P(A)P(O∣A)
So P(A and O)=.522
Question 2 b) The probability of a plant going to seed is 34%. The
probability of that same type of plant surviving the winter is 38%,
and the probability of both is 10%. What is the probability that a
randomly selected plant will go to seed or survive the winter?
The sample space is the set of all plants of that type.
Let S be the event that a plant will seed, and W the event that a plant will survive the winter. We are given P(S)=.34, P(W)=.38, and P(S and W)=.1, and we are asked to find P(S or W). We will use the formula:
P(S or W)=P(S)+P(W)−P(S and W)
Question 3: A hair salon completed a survey of 347 customers about
satisfaction with service and type of customer. A walk-in customer is
one who has seen no ads and not been referred. The other customers
either saw a TV ad or were referred to the salon (but not both). The
Assume the table represents the entire population of customers.
Find the probability that a customer is
Answer: From the table we see that P(Dissatisfied)=34730
b) Dissatisfied and a walk-in
Answer: From the table we see that P(Dissatisfied and a walk-in)=34721
d) Very satisfied, given referred
Answer: There are 143 referred customers, 44 of which are very satisfied. So
P(Very Satisfied, given Referred)=14344
e) Very satisfied or saw a TV ad
Answer: We will use the formula:
P(Very Satisfied or Saw a TV ad)=P(Very Satisfied)+P(Saw a TV ad)−P(Very Satisfied and Saw a TV ad)
From the table we get:
Question 4: A basketball player makes 70% of the free throws he
shoots. Suppose that he tries 15 free throws.
a) What is the probability that he will make more than 7 throws?
b) Find the expected value.
c) Find the standard deviation.
We have a binomial distribution with probability of success (makes the free throw) p=.7, and n=15.
a) We are asked for P(r>7)=P(r=8)+P(r=9)+⋯+P(r=15)
Alternatively we could have used the cumulative probability for the binomial distribution:
b) The expected value is given by μ=np
c) The standard deviation is given by the formula: σ=npq where, in our case, the probability of failure q=1−.7=.3
Question 5: Let x be a random variable that represents the length of
time it takes a student to write a response paper. It was found that
x has an approximately normal distribution with mean μ=7.2
hours and standard deviation σ=1.8 hours.
a) What is the probability that it takes at least 5 hours
for a student to write a response paper?
b) Suppose 20 students are selected at random. What is the
probability that the mean time xˉ of writing a paper for
these 20 students is not more than 8 hours?
Answer: We have a normal distribution with μ=7.2 and σ=1.8, and for Part a) we are asked to find P(x≥5).
b) The distribution of the sample means xˉ is normal with μxˉ=7.2 and standard deviation σxˉ=201.8.
We want the probability P(xˉ≤8)
Question 6: A random sample of 14 candy store franchises had a mean
start up cost of xˉ=$104.70 thousand and s=$28.30
thousand. Find a 95% confidence interval for the population
average start-up cost μ for candy store franchises. Assume x has
a distribution that is approximately normal.
Answer: We have to find a confidence interval using the Student t-distribution. The confidence level is 95% so the sum of the two tails will be α=.05, and therefore each tail should be .25, the area up to the critical value then should be .975. We use the inverse t function, with n−1=13 degrees of freedom:
So we can calculate the margin of error using the formula:
The endpoints of the 95% confidence interval are then xˉ±E=104.70±16.3399
So the interval is [88.36,121.04]
Question 7: Let x be a random variable that represents the
hemoglobin count (HC) in human blood (measured in grams per
milliliter). In healthy adult females, x has an approximately
normal distribution with a population mean of μ=14.2, and
population standard deviation of σ=2.5. Suppose a female
patient had 10 blood tests over the past year, and the sample mean
HC was determined to be x=15.1. With a level of
significance α=.05, determine whether the patient’s HC is
higher than the population average. Specifically, do the following:
a) State the null hypothesis H0 and the alternate hypothesis H1.
b) Determine the value of the sample test statistic (either z or t).
The population standard deviation is known so we are going to use a z-test. This is a right tailed test with level of significance α=.05. The critical z-value is given by the formula:
The p-value for this test is then P(z>1.96)=P(z<−1.96).
Since the p-value is more than the level of significance we fail to rejectH0.
Question 8: In South Africa the size of locust populations may be
related to the average temperature during the time of year when most
insect eggs incubate. In the following table x is a random variable
representing the average temperature over the incubation period in
degrees Celsius while y represents the length of incubation period
a) Plot a scatter diagram of the data
b) Based on a scatter diagram, would you estimate the correlation coefficient to be positive, close to zero, or negative?
Answer: The correlation coefficient should be positive.
c) Interpret your results from parts (a) and (b).
Answer: We expect that the higher the temperature during the incubation period the longer the incubation period is.