Question

If possible, could you write out the process of these problems? include a drawn normal distribution with shading please.

1. suppose X-N (4, 2) what value of x is 1.5 standard deviations to the left of the mean?
2. Suppose X-N (15,3) between what x values does 68.27% of the data lie? The range of x values is centered at the mean the distribution (i.e., 15)
3. X-N (54,8) find the probability that x>56

4. height and weight are two measurements used to track a child's development. The World Health Organization measures a child's development by comparing the weights of children who are the same height and same gender. In 2009 Weights For All 80cm girls in the reference population at a mean of 10.2 kg and a standard deviation of 0.8 kg. X-N (10.2, 0.8). Weights are normally distributed. calculate the Z scores that correspond to the following weights and interpret them.
a. 11 kg b. 7.9kg c. 12.2 kg

5. Facebook provides a variety of statistics on this website that detailed growth in popularity of the site. On average, 28% of 18 to 35 year olds check their Facebook profile before getting out of bed in the morning. Suppose this percentage follows a normal distribution with standard deviation of 5%. A. find the probability that the percent of 18 to 35 year olds to check Facebook before getting out of bed in the morning is at least 30. B. Find the 95th percentile and express it in a sentence.

Answer 1

1) X ~ N (4, 2)

Mean () = 4, standard deviation () = 2

X = - Z

X = 4 - 2*1.5

X = 1

2) X ~ N (15, 3)

Mean () = 15, standard deviation () = 3

- < X < +

15-3 < X < 15+3

12 < X < 18

68.27% data values lies within 12 and 18 in this interval.

3)

X ~ N (54, 8)

Mean () = 54, standard deviation () = 8

Find: P(X > 56)

4) X ~ (10.2, 0.8)

Mean () = 10.2, standard deviation () = 0.8

a) Z = (X - ) /  

Z = (11 - 10.2) / 0.8

Z = 1

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

b)

Z = (X - ) /  

Z = (7.9 - 10.2) / 0.8

Z = -2.875

The weight 7.9 kg is 2.875 standard deviation below the average weight value of 10.2 kg

C)

Z = (X - ) /  

Z = (12.2 - 10.2) / 0.8

Z = 2.5

The weight 12.2 kg is 2.5 standard deviation above the average weight value of 10.2 kg.