STAT Homework2 complete solutions correct answers key

STAT Homework2 complete solutions correct answers key

1.  Chapter 7 Exercise 9. (pg.269) A variable is normally distributed with a mean of 120 and a standard deviation of 5. One score is randomly sampled. What is the probability it is above 127?

P(X>127 | Mu = 120, s=5)=

1-pnorm(127,120,5)

[1] 0.08075666

2.  Chapter 7 Exercise 11. (pg. 269) A group of students at a school takes a history test. The distribution is normal with a mean of 25, and a standard deviation of 4. (a) Everyone who scores in the top 30% of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate? (b) The top 5% of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?

mean score m= 25; s = 4

Using the normal calculator (Ch-7, Lane),

a. the lowest score someone can get and still earn a certificate = 27.096 (or approx.. 27.1)

b. the lowest score someone can get and still go onto compete with the rest of the state = 31.58 (or approx. 31.6)

3. Chapter 7 Exercise 12. (pg. 269) Use the normal distribution to approximate the binomial distribution and find the probability of getting 15 to 18 heads out of 25 flips. Compare this to what you get when you calculate the probability using the binomial distribution. Write your answers out to four decimal places.

For the binomial distribution of coin flipping, probability of getting a head (success) in one flip = p = 0.5

N = no. of trials = 25

Mean is m = N*p = 25 * 0.5 = 12.5              

Variance σ² = Np(1-p) = (25)(0.5)(0.5) = 6.25

The standard deviation is therefore σ = √6.25 = 2.5.

Using the binomial calculator (Lane CH-5),

the probability of getting 15 to 18 heads out of 25 flips = 0.2049

Using the normal distribution to approximate the binomial distribution: The area between 14.5 and 18.5 in the normal distribution curve is an approximation of the probability of obtaining 15 to 18 heads.

Area below 14.5 = 0.7881

Area below 18.5 = 0.9918

Therefore,

Area between 14.5 and 18.5 = 0.9918 – 0.7881 = 0.2037

So in this case,

the probability of getting 15 to 18 heads out of 25 flips = 0.2037

Note: Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

4.  Chapter 6 (Illowsky et al.) Exercise 60. (pg. 363) What is the median recovery time?

a. 2.7

b. 5.3   (In the normal distribution, mean and median are the same

c. 7.4

d. 2.1

5.  Chapter 6 (Illowsky et al.) Exercise 66. (pg. 364) Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean μ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them.

1.      11 kg

2.      7.9 kg

3.      12.2 kg

Answer:

1. z-score for 11 kg = (11- μ)/ σ = (11- 10.2)/ 0.8 = 0.8/0.8 = 1

Interpretation: The weight 11 kg is 1 standard deviation above the average weight value of 10.2 kg.

2. z-score for 7.9 kg = (7.9 - μ)/ σ = (7.9 - 10.2)/ 0.8 = -2.3/0.8 = -2.875

Interpretation: The weight 7.9 kg is 2.875 standard deviations below the average weight value of 10.2 kg.

3. z-score for 12.2 kg = (12.2- μ)/ σ = (12.2 - 10.2)/ 0.8 = 2.0/0.8 = 2.5

Interpretation: The weight 12.2 kg is 2.5 standard deviations above the average weight value of 10.2 kg. As the z-score for this weight is greater than 2, it can be concluded that 12.2 kg weight for a girl 8 cm tall is an “unusual” value.