Data Science with R

Discussion in 'General Discussions' started by Indu Voleti, Nov 24, 2019.

  1. Indu Voleti

    Indu Voleti Member

    Joined:
    Sep 8, 2019
    Messages:
    2
    Likes Received:
    0
    # For a given Mean 494 and standard deviation = 100
    # 1. What is the probability that a randomly selected score
    # is between 600 and its mean.
    # 2. What is the probability of obtaining a score more than 700.
    # 3. Score that is less than 550.
    # 4. Score between 300 and 600.
    pnorm(600,494,100)-pnorm(494,494,100)
    #OUTPUT:
    # > pnorm(600,494,100)-pnorm(494,494,100)
    # [1] 0.3554277
    1-pnorm(700,494,100)
    # OUTPUT:
    # > 1-pnorm(700,494,100)
    # [1] 0.01969927
    pnorm(550,494,100)
    #OUTPUT:
    # > pnorm(550,494,100)
    # [1] 0.7122603
    pnorm(600,494,100)-pnorm(300,494,100)
    #OUTPUT:
    # > pnorm(600,494,100)-pnorm(300,494,100)
    # [1] 0.8292379
    #2.Suppose during any hour in large departmental store, the average number of shoppers is 448,
    # with the standard deviation of 21 shoppers. What is the probability that a random sample of 49
    # different shopping hours will yield a sample mean between 441 and 446 shoppers.
    Sample_mean <- 448
    Sample_sd <- 21
    Sample_size <- 49
    pnorm(446,Sample_mean,Sample_sd/sqrt(Sample_size))-pnorm(441,Sample_mean,Sample_sd/sqrt(Sample_size))
    # OUTPUT:
    # > pnorm(446,Sample_mean,Sample_sd/sqrt(Sample_size))-pnorm(441,Sample_mean,Sample_sd/sqrt(Sample_size))
    # [1] 0.2426772
    # 3.Mercury makes a 2.4 lt V-6 engine, The Laser XRi, used in speedboats. The companies engineer believe
    # that the engine delivers an average power of 220 horsepower and that the standard deviation of power
    # delivered is 15 horsepower. A potential buyer intends to sample 100 engines(each engine to be run a
    # single time). What is the probability that the sample mean will be less than 217 horsepower?
    Avg_power <- 220
    Sample_sd <- 15
    Sample_size <- 100
    pnorm(217,Avg_power,Sample_sd/sqrt(Sample_size))
    # OUTPUT:
    # > pnorm(217,Avg_power,Sample_sd/sqrt(Sample_size))
    # [1] 0.02275013
    #4.Comcast, the computer services company, is planning to invest heavily in online television services. As
    # part of the decision, the company wants to estimate the average no of online shows a family of four would
    # watch per day. A random sample of n=100 families is obtained, and in this sample the average no of shows
    # viewed per day is 6.5 and the population standard deviation is known to be 3.2. Construct a 95%
    # confidence interval for the average no of online television shows watched by the entire population of
    # families of four.
    Sample_mean <- 6.5
    Pop_Sd <- 3.2
    Sample_size <- 100
    CI <- 0.95
    z <- qnorm(0.975)
    mean_1 <- Sample_mean + z*(Pop_Sd/sqrt(Sample_size))
    mean_2 <- Sample_mean - z*(Pop_Sd/sqrt(Sample_size))
    mean_1
    mean_2
    # OUTPUT:
    # > Sample_mean <- 6.5
    # > Pop_Sd <- 3.2
    # > Sample_size <- 100
    # > CI <- 0.95
    # >
    # > z <- qnorm(0.975)
    # > mean_1 <- Sample_mean + z*(Pop_Sd/sqrt(Sample_size))
    # > mean_2 <- Sample_mean - z*(Pop_Sd/sqrt(Sample_size))
    # > mean_1
    # [1] 7.127188
    # > mean_2
    # [1] 5.872812
    # 5. A stock market analyst wants to estimate the average return on a certain stock. A random sample of 15
    # days yields an average (annualized) return of Xbar=10.37% and a standard deviation of s=3.5%. Assuming
    # a normal population of returns, give a 95% confidence interval for the average return on this stock.
    Sample_size <- 15
    Sample_mean <- 10.37
    Sample_sd <- 3.5
    CI <-0.95
    # Here Pop Standard Deviation is not know so we cannot use Z-Distrubution
    # So we use T-Distrubution
    t <- qt(0.975,Sample_size-1)
    Pop_Mean1 <- Sample_mean + t*(Sample_sd /sqrt(Sample_size))
    Pop_Mean2 <- Sample_mean - t*(Sample_sd /sqrt(Sample_size))
    Pop_Mean1
    Pop_Mean2
    # OUTPUT:
    # > Pop_Mean1
    # [1] 12.30824
    # > Pop_Mean2
    # [1] 8.431765
     
    #1

Share This Page