Download 1. Consider a binomial distribution with 15 identical trials, and a

1. Consider a binomial distribution with 15 identical trials, and a probability of success of 0.5 i.
Find the probability that x = 3 using the binomial tables ii. Use the normal approximation to find
the probability that x = 3. Show all work.
i. 0.01389
ii.
a. Mean = np = 15 * 0.5 = 7.5
b. Sd = sqrt[np(1-p)] = sqrt[15 * 0.5 * 0.5] = sqrt(3.75) = 1.936
z = (x - mu)/sigma
z = (2.5 - 7.5)/1.936
z = -5/1.936
z =-2.58
p = 0.0049
z = (x - mu)/sigma
z = (3.5 - 7.5)/1.936
z = -4/1.936
z =-2.07
p = 0.0194
0.0194 - 0.0049 = 0.0145
2. The diameters of oranges in a certain orchard are normally distributed with a mean of 7.25
inches and a standard deviation of 0.50 inches. Show all work. (A) What percentage of the
oranges in this orchard have diameters less than 6.8 inches?
z = (x - mu)/sigma
z = (6.8 - 7.25)/0.25
z = -0.45/0.25
z =-1.8
p = 0.0359
3.6%
(B) What percentage of the oranges in this orchard are larger than 7.10 inches?
z = (x - mu)/sigma
z = (7.1 - 7.25)/0.25
z = -0.15/0.25
z =-0.600000000000001
p = 0.2743
1 – 0.2743 = 0.7257
72.6%
(C) A random sample of 100 oranges is gathered and the mean diameter obtained was 7.10. If
another sample of 100 is taken, what is the probability that its sample mean will be greater
than 7.10 inches?
z = (x - mu)/(sigma/√n)
z = (7.1 - 7.1)/(0.25/√100)
z = 0 / 0.025
z =0
p = 0.5
50%
(D) Why is the z-score used in answering (A), (B), and (C)?
Because we need to normalize the score based on the standard deviation compared to the
mean.
(E) Why is the formula for z used in (C) different from that used in (A) and (B)?
B ecause in C we’re looking at a sample, not a single value.
3. Assume that the population of heights of female college students is approximately normally
distributed with mean m of 70 inches and standard deviation s of 3.75 inches. A random sample
of 16 heights is obtained. Show all work.
(A) Find the proportion of female college students whose height is greater than 68 inches.
z = (x - mu)/sigma
z = (68 - 70)/3.75
z = -2/3.75
z =-0.533333333333333
p = 0.296901
BUT since we want "greater than", we need to subtract the probability from 1
p = 1 - 0.2969 = 0.7031
(B) Find the mean and standard error of the distribution
Mean = 70
SE = 3.75/sqrt(16) = 0.9375
(C) Find p(x with a line over it, > 68)
z = (x - mu)/(sigma/√n)
z = (68 - 70)/(3.75/√16)
z = -2 / 0.9375
z =-2.13
p = 0.0164
BUT since we want "greater than", we need to subtract the probability from 1
p = 1 - 0.0164 = 0.9836
4. Answer the following questions regarding the normal, standard normal and binomial
distributions. (A) Why can the normal distribution be used as an approximation to the binomial
distribution?
Because the binomial distribution approximates the normal distribution as n increases. The
discrete probability distribution resembles a continuous probability distribution.
(B) What conditions must be met in order to use the normal distribution to approximate the
binomial distribution?
np>10 and np(1-p)>10
(C) What are the advantages of using the standard normal distribution over the normal
distribution?
Because the z score is found by using the standard normal distribution. This represents the
number of standard deviations away from the mean that the score is.
5. Two yellow socks are selected, one at a time from a clothes drawer containing 6 black, 6
brown and 6 green socks. Let x represent the number of yellow socks selected in 2 selections
from the drawer. (A) If this experiment is completed without replacing the socks each time,
explain why x is not a binomial random variable.
Because the trials are not independent. The probabilities change each time.
(B) If this experiment is completed with replacement of the socks each time, explain why x is a
binomial random variable.
Because the trials are independent of one another. Each trial is not affected by the previous
one. There is the same chance of success on each trial. There is a fixed number of trials. The
only two possible outcomes are success and failure.