Heights of 10 year olds, regardless of gender, closely follow a normal distribut

Question

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.


(a) What is the probability that a randomly chosen 10 year old is shorter than 48 inches?
(please round to four decimal places)


(b) What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?
(please round to four decimal places)


(c) If the tallest 10% of the class is considered "very tall", what is the height cutoff for "very tall"?
inches (please round to two decimal places)


(d) The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What proportion of 10 year olds cannot go on this ride?
(please round to four decimal places)

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    48      
u = mean =    55      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    -1.166666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.166666667   ) =    0.1217 [ANSWER]

**************

b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    60      
x2 = upper bound =    65      
u = mean =    55      
          
s = standard deviation =    6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.833333333      
z2 = upper z score = (x2 - u) / s =    1.666666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.797671619      
P(z < z2) =    0.952209648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.1545 [ANSWER]

*******************

c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    55      
z = the critical z score =    1.281551566      
s = standard deviation =    6      
          
Then          
          
x = critical value =    62.69 [ANSWER]

********************

d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    54      
u = mean =    55      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    -0.166666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.166666667   ) =    0.4338 [ANSWER]

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