Statistical Distributions (1) Flashcards
The probability function for the discrete random variable X is given by P(X = x) = (1/k) x^2 for x = 1,2,3,4
Find the value of k and P(X <= 2)
As probabilities add up to 1 (1/k) 1^2 + (1/k) 2^2 + (1/k) 3^2 + (1/k) 4^2 = 1 (1/k)(1 + 4 + 9 + 16) = 1 30/k = 1 k = 30
P(X <= 2) = P(X = 1) + P(X = 2)
= (1/30)(1^2) + (1/30)(2^2) = 5/30 = 1/6
The random variable X follows the binomial distribution X ~ B(12,0.6). Find:
P(X < 8)
P(X = 5)
P(3 < X <= 7)
Binomial CD: P(X<8) = P(X<=7) = 0.5618 (4 d.p.)
Binomial PD:
P(X=5) = 0.1009 (4 d.p.)
P(3 < X <= 7) = P(X <= 7) - P(X <= 3) = 0.5466
Apples are stored in crates of 40. The probability of any apple containing a maggot is 0.15, and is independent of any other apple containing a maggot. In a random sample of 40 apples, find the probability that:
Fewer than 6 apples contain maggots
Let X represent the number of apples that contain a maggot
Then X ~ B(40, 0.15). So P(X < 6) = P(X <= 5) = 0.4325
Apples are stored in crates of 40. The probability of any apple containing a maggot is 0.15, and is independent of any other apple containing a maggot. In a random sample of 40 apples, find the probability that:
More than 2 apples contain maggots
P(X > 2) = 1 - P(X <= 2) = 0.9514
Apples are stored in crates of 40. The probability of any apple containing a maggot is 0.15, and is independent of any other apple containing a maggot. In a random sample of 40 apples, find the probability that:
Jin has 3 crates. Find the probability that more than 1 crate contains more than 2 apples with maggots
The probability that a crate contains more than 2 apples with maggots is 0.9514.
So define a random variable Y, where Y is the number of crates that contain more than 2 apples with maggots. Then Y ~ B(3, 0.9514)
P(Y > 1) = 1 - P(Y <= 1) = 0.9931
Apples are stored in crates of 40. The probability of any apple containing a maggot is 0.15, and is independent of any other apple containing a maggot. In a random sample of 40 apples, find the probability that:
Give one criticism of the assumption that apples contain maggots independently of each other
Maggots could spread to other nearby apples in a crate, some trees might have more apples with maggots than others
The random variable X has a normal distribution with mean 120 and standard deviation 25.
Find P(X > 145)
Find the value of j such that P(120 < X < j) = 0.4641
LB: 145 UB: 9999 σ = 25 μ = 120 Using normal CDF function: P(X > 145) = 0.159
P(X < j) - P(X <= 120) = 0.4641
P(X < j) = 0.4641 + P(X <= 120) = 0.4641 + 0.5 = 0.9461
Using the inverse normal function: j = 165
A garden centre sells bags of compost. The volume of compost in the bags is normally distributed with a mean of 50 litres
If the standard deviation of the volume is 0.4 litres, find the probability that a randomly selected bag will contain less than 49 litres of compost
Let X be the volume of compost in a bag, so X ~ N(50, 0.4^2). LB: -9999 UB: 49 σ = 0.4 μ = 50
Using the normal CDF: P(X < 49) = 0.0062
A garden centre sells bags of compost. The volume of compost in the bags is normally distributed with a mean of 50 litres
If 1000 of these bags of compost are bought, how many bags would you expect to contain more than 50.5 litres of compost?
LB: 50.5
UB: 9999
σ = 0.4
μ = 50
Using the normal CDF: P(X > 50.5) = 0.1056…
So in 1000 bags, 0.1056… * 1000 = 106 bags approximately
A sweet shop sells giant marshmellows. The mass of a marshmellow, in grams, is described by the random variable Y, where Y ~ N(75, σ^2). It is found that 10% of the marshmellows weigh less than 74 grams. Find σ
Convert Y to the standard normal distribution Z
P(Y < 74) = 0.1 means P(Z < (74-75)/σ) = P(Z < -1/σ) = 0.1
Calculate this
The lifetimes of a particular type of battery are normally distributed with mean μ hours and standard deviation σ hours. A student using these batteries finds that 40% last less than 20 hours and 80% last less than 30 hours. Find σ and μ
Let X be the lifetime of a battery, so X ~ N(μ, σ^2)
Then P(X < 20) = 0.4 and P(X < 30) = 0.8
Convert X to the standard normal distribution Z
P(Z < (20 - μ)/σ)) = 0.4
P(Z < (30 - μ)/σ)) = 0.8
Calculate this to get -0.2533 and 0.8416
Rewrite these to get: 30 - μ = 0.8416σ 20 - μ = -0.2533σ Subtract these two to get: 10 = (0.8416 + 0.2533)σ
Find σ and then sub the solution to σ into 20 - μ = -0.2533σ
The random variable X is binomially distributed with X ~ B(100,0.6).
State the conditions needed for X to be well approximated by a normal distribution
Explain why a continuity correction is necessary in these circumstances
Need n to be large and p to be close to 0.5
A binomial distribution is discrete, whereas a normal distribution is continuous. The continuity correction means probabilities
The random variable X is binomially distributed with X ~ B(100,0.6).
Using a suitable approximation, find:
P(X >= 65)
P(50 < X < 62)
n is large and p is fairly close to 0.5, so use the normal approximation N(60,24)
P(X >= 65) = P(Y > 64.5) = 0.179
P(50 < X < 62) = P(Y < 61.5) - P(Y < 50.5) = 0.594
On average, 55% of customers in a cafe order a cup of coffee. One day, the cafe served 150 customers. At the start of the day, the manager calculated that they had enough coffee to make 75 cups of coffee. By using a suitable approximation, find the probability that more than 75 customers try to order coffee
Let X represent the number of cups of coffee sold
So X ~ B(150,0.55)
np = 150 * 0.55 = 82.5 np(1-p) = 82.5 * 0.45 = 37.125
So use the normal approximation Y ~ N(82.5,37.125)
LB: 75.5
UB: 9999
σ = ¬37.125 = 6.093028….
μ = 82.50
P(X > 75) = P(Y > 75.5) = 0.875 (3 s.f.)
The random variable follows a binomial distribution of X ~ B(n,p). X is approximated by a normally distributed variable Y, with a standard deviation of 6. Using this normal approximation P(X > 127) = 0.9970
Find the mean of the normal approximation
Hence estimate n and p
The normal approximation is Y ~ N(μ, 6^2)
P(X > 127) = P(X > 127.5)
= P(Z > ((127.5 - μ)/6)) = 0.9970
Using the inverse normal function
((127.5 - μ)/6) = -2.747…
127.5 - μ = -16.486…
μ = 144 to 3 s.f.
μ = np np = 143.986... 6^2 = np(1-p) np(1-p) = 36
Dividing 2nd equation by 1st equation:
1 - p = 36 / 143.986… = 0.25002…
p = 0.74997… = 0.750 (3 s.f.)
Then n = 143.986… / 0.74997. = 192 (3 s.f.)
