Question

1.Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. What is the probability that a randomly chosen 10 year old is shorter than 48 inches?

2. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?

3.Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. If the tallest 10% of the class is considered “very tall”, what is the height cutoff for “very tall”?

4.Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

Answer 1

Solution (1): Let X be the heights of 10-year olds. Here, X is normally distributed. Let the mean height be u= 55 inches. Let the standard deviation be o = 6 inches. The probability that a randomly selected chosen 10-year old is shorter than 48 inches is, P(X < 48)= P( 4-4-48–55) 06 = P(Z <-1.67) = 0.1210 From standard distribution tables/

Solution (2): Let X be the heights of 10-year olds. Here, X is normally distributed. Let the mean height be u= 55 inches. Let the standard deviation be o = 6 inches. The probability that a randomly selected chosen 10-year old is between 60 and 65 inches *565) P(60**.43.09.25 IS- 60-55 X-u 65-55 P(60 SX S65) =P 6 0 6 = P(0.83 52 51.67) = P(251.67)- P(2 50.83) = 0.9525 -0.7967 = 0.1558 [From standard distribution tables

Solution (3): Let X be the heights of 10-year olds. Here, X is normally distributed. Let the mean height be u= 55 inches. Let the standard deviation beo = 6 inches. For bottom 10% the equation becomes, P(X<X)= 0.10 p(4544*6*55) -0.10 Using the standard normal distribution area tables, the Z-score, corresponding to an cumulative area of 0.10 is -1.282. So, the equation becomes, P(Z <-1.282)=0.10 Comparing (1) and (2), xa -55 = -1.282 40 = 55-(1.282)6) xa = 47.31

Solution (4): Let X be the heights of 10-year olds. Here, X is normally distributed. Let the mean height be u = 55 inches. Let the standard deviation beo = 6 inches. If the height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches, then the probability that a randomly selected chosen 10-year old is cannot go on this ride is. P(X <54) = P(454_5455) = P(Z <-0.17) = 0.4325 [From standard distribution tables] = 43.25%