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Question 1
1. Z= x-μ/standard deviation s
A) Given the following information:
X= 37
Mean= 54 years
Standard deviation= 11 years
Z= (x- μ)/standard deviation
Z= -1.
Hence the z score corresponding to X=37 is -1.
B) We can use the formula for probability
P( z< -1.545)
Using the z table we find at z= -1.54 we have a value of 0.0618 and for z= -1.55 the value is
Hence since our z score is -1.545 computed in part a, we take average of our critical values
P( X <37) = 0.
The number of voters voting in the 2020 general election has a 6.11 % chance of being less than
37 years old
C) Since we are finding the number of people aged more than 37 years old, we use the
following;
P( X> 37) = 1- P(X< 37)
The number of voters who voted in the 2020 general election has a 93.89% chance of being more
than 37 years old.
Question 2
a)
0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.
-1.
-0. 0
1
2
Probability Plot
b) Correlation coefficient between the x and the corresponding z scores is 0.
c) At the value assigned of 16 and the critical value found to be 0.941, we find the following: Critical value < Correlation coefficient 0.94 < 0. Since the critical value is less than correlation coefficient, it is reasonable to assume that the population is normally distributed d) Since the critical value is less than the correlation coefficient, the ER wait times are normally distributed and hence the mean=median= mode. Patients are able to go to the emergency rooms into the exact time as no patient has been seeing to be sent to the emergency room too ealy or too late. Each patient has been admitted to the emergency room with the exact wait time and the response time by the Hospital to be exact. Question 3 a) We would have to calculate the probability using the following: Given that mean= 287 Standard deviation= 15 days P( x< 280)= (x- mean)/standard deviation Z score = (280- 287)/ = -0. P (Z <-0.46)
Z score
X values
The prescription drug prices for 10 individuals sampled is $7. Question 5 A) Proportion = number of successes/ sample size Given that 245 students said they use social media and 350 students are sampled, Proportion of students who use socials media= 245/ = 0. The sample proportion of teenagers aged 13 to 17 years of age who use social media is 70%. B) P (pS < 0.70)= pS – population proportion/ square root of 0.77 (1-p)/ = ps-p/square root of p (1-p)/sample size. The sample proportion is 0.70 (number of successes) and the population proportion p is 0.77, sample size is 350. = 0.70-0.77/ square root of 0.77(1-0.77)/ = -0.7/0. = -3. (P <-3.11)= 0. = 0. We can say that 0.09% of 350 students would have a sample proportion lower than that of the 350 students sampled. Question 6 a) Number of murders committed with a firearm= % murders * sample size =0.712* =284. If 400 murders are randomly selected, approximately 284.8 murders are committed with a firearm b) We will be using the 95% interval with 2 standard deviations apart p=0.712 murders committed q= 0.288 murders not committed q=1-p Standard deviation = square root of np (1-p) = 400 * 0.712 * 0. = 82. Standard deviation = 9. (lower bound, upper bound) (284.8-2(9.05661887)), (284.8+2(9.05661887) = (266.68, 302.9) Yes 316 murders by firearm is unusual in a random sample of 400 murders because 316 murders falls outside our range with a 95% interval of (266.68, 302.9).
Question 7
a) We are given the following:
Mean= 9.5 minutes
Standard deviation= 1.25 minutes
First we need to find the probability that it will take less than 10 minutes
P (x <10)= (x- mean)/ standard deviation
Then we will find the probability to find the time it takes more than 10 minutes
P (x > 10)= 1- P (x <10)
The ambulance callouts have a 34.46% chance of incurring a financial penalty for the hospital if
it takes more than 10 minutes.
b) We are provided the following
Mean= 9.5 minutes
Standard deviation= 1.25 minutes
We are also provided the probability being 5%
Locating 5% in the z table is between -2.57 and -2.58 which have values of 0.0051 and 0.
respectively
0.05 is between the 2 z values which gives us -2.
Z=-2.
X= mean+z (standard deviation)
There is a 6.28% chance that the ambulance call outs with mean of 9.5 minutes and standard
deviation of 1.25 minutes is within the guaranteed time limit.
P(x=5) = (12!/5!) * (0.267)^5 * (1-0.267)12- = 792 * 0.001356926 * 0. = 0. P(x=6) = (12!/6!) * (0.267)^6 * (1-0.267)12- = 924 * 0.000362299 * 0. = 0. P(X=7)= (12!/7!) * (0.267)^7 * (1-0.267)12- = 792 * 0.000096734 * 0. = 0. P(X<8)= P(x=0) + P(x=1) +P(x=2) +P(x=3) +P(x=4) +P(x=5) +P(x=6)+ P(X=7) = 0. For those greater than 8 individuals, P (X>8)= 1-0. = 0. = 0.43% There is a 0.43% chance that among 12 randomly observed individuals greater than 8 has a do not cover their mouth when sneezing
b) P= 0.267 (the probability of those people not covering their mouth when sneezing)
q=1-p
P(X<3)= P(x=0)+P(x=1)+P(x=2) P=0 = (12!/0!) * (0.267)^0 * (1-0.267)12- = 0. P=1. = (12!/1!) * (0.267)^1 * (1-0.267)12-
P=2. = (12!/2!) * (0.267)^2 * (1-0.267)12-
There is a 21% chance that among 12 randomly observed individuals fewer than 3 do not cover their mouth when sneezing. c) First we will find the probability of the 12 individuals greater than half who have covered their mouth when sneezing.
P= 0.267 (the probability of those people not covering their mouth when sneezing)
q=1-p
P(X>5)= P(x=0) + P(x=1) +P(x=2) +P(x=3) +P(x=4) +P(x=5) +P(x=6) P(x=0) = (12!/0!) * (0.267)^0 * (1-0.267)12- = 0. P(x=1)= (12!/1!) * (0.267)^1 * (1-0.267)12- = 12 * 0.267 * 0. = 0. P(x=2) = (12!/2!) * 0.267^2 * (1-0.267)12- = 66 * 0.071289 * 0. = 0. P(x=3) = (12!/3!) * 0.267^3 * (1-0.267)12- = 220 * 0.019034163 * 0. = 0.
b) Based on analysis, Program B is better due to the standard deviation being 10.17 pounds while
the standard deviation for Program A was 11.04 pounds was greater. With the standard deviation
being less, more people would have produced better results in terms of weight loss program as
the data would be closer to the mean with the reduced variation.
c) Some information we may need to consider is the age group with the results produced, clearly
different people of different ages would produce a different set of results. The type of workouts
done need to be defined whether they were based on cardio only or weight lifting. Also, the diet
plan of individuals need to be described in terms of the meal plan or the keto diet being followed.
Also, the effect of supplements taken by individuals need to be considered which might have
produced a loss of pounds for people.
