BUS 270: Statistics Assignment on Probability and Distributions

This statistics assignment provides solutions to problems involving various probability distributions. The assignment covers topics such as identifying continuous and discrete variables, applying the binomial distribution to calculate probabilities and expected values, and utilizing the normal distribution to determine probabilities and percentiles. It includes calculations for standard error and mean, and explores probability distributions, including expected value and standard deviation. The assignment also validates probability distributions based on given conditions. The document also provides a solution for an extra credit question involving the estimation of defective items based on a sample proportion.

Statistics Assignment
[Document subtitle]
[DATE]

Question 6
a) Continuous
b) Continuous
c) Discrete
d) Continuous
Question 7
Binomial distribution
Defective gas tanks Probability of success (p) = 0.40
Sample (n) = 11
(a) Probability that exactly 5 of gas tank would be defective
P (5) = 0.2207
(b) Expected number of gas tank
1

Probability of being defective tank = 0.40
Probability of being non defective tank = 1- 0.40 = 0.60
Expected number of gas tank = np = 11*0.60 = 66
Question 8
Normal distribution
Mean = 6 ounces
Standard deviation = 0.3 ounces
(a) % of bottles contains greater than 6.37 ounces of vitamin
10.93% of bottles contains greater than 6.37 ounces of vitamin.
(b) Number of bottles of vitamin = 40
Probability that bottle would contain greater than 6.51 ounces of vitamin
Standard error = standard deviation/ sqrt (sample size) = 0.3 / sqrt (40) = 0.04743
2

(c) Number of ounces in top 90% of bottles
P(Z<z) = 1.28
Let number of ounces is X.
Hence,
X = 6 + (1.28*0.3) = 6.384
Number of ounces of vitamin in top 90% of bottles would be 6.49356.384.
Question 9
Normal distribution
Mean = 4.74 cars
Standard deviation = 4 cars
Sample size = 81
3

(a) Standard error
Standard error = standard deviation / sqrt (sample size) = 4 / sqrt (81) = 0.444
(b) Probability that mean would be between 3.5 and 4.9 cars
Question 10
x 1 2
p(x) 1/4 3/4
x. p(x) 0.25 1.5
(x-mean)^2 * p(x) 0.140 0.0468
Probability distribution
Mean = (0.25) + (1.5) = 1.75
Standard deviation = sqrt (0.140+0.0468) = 0.432
4

Question 11
(a) The probability distribution is valid because the given probabilities are non-negative and
sum of distribution is equal to 1. The two conditions required here is highlighted below.
The sum of all probabilities of probability distribution must be 1.
Probabilities must be positive.
(b) P(X>=2)
P(X>=2) = 0.13 + 0.05 + 0.43 = 0.61
(c) Number of cheesecakes sells
x p(x) x p(x)
0 0.16 0
1 0.23 0.23
2 0.13 0.26
3 0.05 0.15
4 0.43 1.72
total 2.36
Expected value = E(x) = ∑ x . p ( x ) =2 .3 6
Question 12 (Extra Credit)
Sample = 400
Number of defective items = 10% of 3100 = 3100*10% = 310
(a) Point estimate of proportion of defective item
P = 0.10
5

(b) Standard error of sample proportion of defective item
Standard erro r = √ p ( 1− p )
n = √0.10 1−0.10
400 =0.015
6