The heights of kindergarten children are approximately normally distributed with
ID: 3134594 • Letter: T
Question
The heights of kindergarten children are approximately normally distributed with the following. (Give your answers correct to four decimal places.)
= 43 and = 3.3 inches
(a) If an individual kindergarten child is selected at random, what is the probability that he or she has a height between 41.8 and 44.2 inches?
(b) A classroom of 22 of these children is used as a sample. What is the probability that the class mean x is between 41.8 and 44.2 inches?
(c) If an individual kindergarten child is selected at random, what is the probability that he or she is taller than 45.2 inches?
(d) A classroom of 22 of these kindergarten children is used as a sample. What is the probability that the class mean x is greater than 45.2 inches?
A popular flashlight that uses two D-size batteries was selected, and several of the same models were purchased to test the "continuous-use life" of D batteries. As fresh batteries were installed, each flashlight was turned on and the time noted. When the flashlight no longer produced light, the time was again noted. The resulting "life" data from Rayovac batteries had a mean of 20.4 hours. Assume these values have a normal distribution with a standard deviation of 1.37 hours. (Give your answers correct to four decimal places.)
(a) What is the probability that one randomly selected Rayovac battery will have a test life between 19.9 and 21.4 hours?
(b) What is the probability that a randomly selected sample of 6 Rayovac batteries will have a mean test life between 19.9 and 21.4 hours?
(c) What is the probability that a randomly selected sample of 16 Rayovac batteries will have a mean test life between 19.9 and 21.4 hours?
(d) What is the probability that a randomly selected sample of 65 Rayovac batteries will have a mean test life between 19.9 and 21.4 hours?
(e) Describe the effect that the increase in sample size had on the answers for parts (b) - (d).As sample size increased, the standard error increased, resulting in higher probabilities.As sample size increased, the standard error increased, resulting in lower probabilities. As sample size increased, the standard error decreased, resulting in higher probabilities.As sample size increased, the standard error decreased, resulting in lower probabilities.
The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean = $635 and standard deviation = $42.
(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Use 4 decimal places.)
(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.06? (Round to the nearest dollar.)
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Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of answers to different questions are independent. Jodi is a good student for whom p = 0.83.
(a) Use the normal approximation to find the probability that Jodi scores 78% or lower on a 100-question test.(Round the probablity to the 4th decimal place,)
(b) If the test contains 250 questions, what is the probability that Jodi will score 78% or lower? (Round the probablity to the 4th decimal place,)
(c) How many questions must the test contain in order to reduce the standard deviation of Jodi's proportion of correct answers to half its value for a 100-item test?
Explanation / Answer
The heights of kindergarten children are approximately normally distributed with the following. (Give your answers correct to four decimal places.)
= 43 and = 3.3 inches
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a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 41.8
x2 = upper bound = 44.2
u = mean = 43
s = standard deviation = 3.3
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.363636364
z2 = upper z score = (x2 - u) / s = 0.363636364
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.358064784
P(z < z2) = 0.641935216
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.283870431 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 41.8
x2 = upper bound = 44.2
u = mean = 43
n = sample size = 22
s = standard deviation = 3.3
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.705605731
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.705605731
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.044040756
P(z < z2) = 0.955959244
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.911918488 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 45.2
u = mean = 43
s = standard deviation = 3.3
Thus,
z = (x - u) / s = 0.666666667
Thus, using a table/technology, the right tailed area of this is
P(z > 0.666666667 ) = 0.252492538 [ANSWER]
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d)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 45.2
u = mean = 43
n = sample size = 22
s = standard deviation = 3.3
Thus,
z = (x - u) * sqrt(n) / s = 3.12694384
Thus, using a table/technology, the right tailed area of this is
P(z > 3.12694384 ) = 0.000883168 [ANSWER]
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