stats week 5

Jamie's points per game of bowling are normally distributed with a standard deviation of 14 points. If Jamie scores 92 points, and the z-score of this value is −4, then what is her mean points in a game?

148 points​ We can work backwards using the z-score formula to find the mean. The problem gives us the values for z, x and σ. So, let's substitute these numbers back into the formula: z−4−56−148148=x−μσ=92−μ14=92−μ=−μ=μ We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=92 is four standard deviations to the left of the mean, and each standard deviation is 14. So four standard deviations is (−4)(14)=−56 points. So, now we know that 92 is 56 units to the left of the mean. (In other words, the mean is 56 units to the right of x=92.) So the mean is 92+56=148.

Floretta's points per basketball game are normally distributed with a standard deviation of 4 points. If Floretta scores 10 points, and the z-score of this value is −4, then what is her mean points in a game?

26 We can work backwards using the z-score formula to find the mean. The problem gives us the values for z, x and σ. So, let's substitute these numbers back into the formula: z−4−16−2626=x−μσ=10−μ4=10−μ=−μ=μ We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=10 is four standard deviations to the left of the mean, and each standard deviation is 4. So four standard deviations is (−4)(4)=−16 points. So, now we know that 10 is 16 units to the left of the mean. (In other words, the mean is 16 units to the right of x=10.) So the mean is 10+16=26.

Alice scores an average of 21 points per basketball game, and her points per basketball game are normally distributed. Suppose Alice scores 5 points in a game, and this value has a z-score of −4. What is the standard deviation? Do not include the units in your answer. For example, if you found that the standard deviation was 10 points, you would enter 10.

4 We can work backwards using the z-score formula to find the standard deviation. The problem gives us the values for z, x and μ. So, let's substitute these numbers back into the formula: z−4−4−4σσ=x−μσ=5−21σ=−16σ=−16=4 We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=5 is four standard deviations to the left of the mean, 21. So we can think of the distance between 5 and 21, which is 16. This distance is broken up into four standard deviations, so 164=4. So 4 is the standard deviation.

Alice was told that her reading test score was 2 standard deviations below the mean. If test scores were approximately normal with μ=84 and σ=8, what was Alice's score?

68 We can work backwards using the z-score formula to find the x-value. The problem gives us the values for z, μ and σ. So, let's substitute these numbers back into the formula: z−2−1668=x−μσ=x−848=x−84=x We can think of this conceptually as well. We know that the z-score is −2, which tells us that x is two standard deviations to the left of the mean, 84. So we can think of the distance between 84 and the x-value as (2)(8)=16. So Alice's score is 84−16=68.

Charles has collected data to find that the total snowfall per year in Reamstown has a normal distribution. Using the Empirical Rule, what is the probability that in a randomly selected year, the snowfall was less than 87 inches if the mean is 72 inches and the standard deviation is 15 inches?

84% Notice that 87 inches is one standard deviation greater than the mean. Based on the Empirical Rule, 68% of the yearly snowfalls are within one standard deviation of the mean. Since the normal distribution is symmetric, this implies that 16% of the yearly snowfalls are greater than one standard deviation above the mean. Alternatively, 84% of the yearly snowfalls are less than one standard deviation above the mean.

After collecting the data, Kenneth finds that the body weights of the forty students in a class are normally distributed with mean 140 pounds and standard deviation 9 pounds. Use the Empirical Rule to find the probability that a randomly selected student has a body weight of greater than 113 pounds. Provide the final answer as a percent rounded to two decimal places.

99.85% Notice that 113 pounds is 3 standard deviations less than the mean. Based on the Empirical Rule, 99.7% of the body weights are within 3 standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the body weights are less than the weight that is 3 standard deviations below the mean. Alternatively, 99.85% of the body weights are greater than the weight that is 3 standard deviations below the mean. Therefore, the probability that a randomly selected student has a body weight of greater than 113 pounds is 99.85%.

The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the largest mean.

A Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. The distribution that is farthest to the right is A, so that has the largest mean.

In a small town, 50% of single family homes have a front porch. 48 single family homes are randomly selected. Let X represent the number of single family homes with a porch. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

N(24, 35) Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq‾‾‾‾√. In symbols, the binomial distribution is approximated by N(np,npq‾‾‾‾√). In this case, p=q=12, and n=48. So we find that a good approximation of the distribution of X is μ=12(48)=24. And σ is σ=npq‾‾‾‾√=(48)(12)(12)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=12‾‾‾√≈3.5 So the best approximation is N(24,3.5).

Jerome averages 16 points a game with a standard deviation of 4 points. Suppose Jerome's points per game are normally distributed. Let X = the number of points per game. Then X∼N(16,4). Suppose Jerome scores 10 points in the game on Monday. The z-score when x=10 is _______. This z-score tells you that x=10 is _______ standard deviations to the _______(right /left) of the mean, _______. Correctly fill in the blanks in the statement above.

Suppose Jerome scores 10 points in the game on Monday. The z-score when x=10 is z=−1.5. This z-score tells you that x=10 is 1.5 standard deviations to the left of the mean, 16." The z-score can be found using this formula: z=10−164=−64=−1.5 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, scoring 10 points is 1.5 standard deviations away from the mean. Negative values of z means that that the points per game is below (or to left of) the mean, which was given in the problem: μ=16 points per game.

Rosetta averages 148 points per bowling game with a standard deviation of 14 points. Suppose Rosetta's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(148,14). If necessary, round to three decimal places.

The z-score can be found using this formula: z=x−μσ=110−14814=−3814≈−2.714 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, scoring 110 points is 2.714 standard deviations away from the mean. A negative value of z means that that the value is below (or to the left of) the mean, which was given in the problem: μ=148 points in the game.

In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches. Suppose X, height in inches of adult women, follows a normal distribution. Let x=68, the height of a woman who is 5' 8" tall. Find and interpret the z-score of the standardized normal random variable.

This means that x=68 is one standard deviation (1σ) above or to the right of the mean, μ=64. X is a normally distributed random variable with μ=64 (mean) and σ=4 (standard deviation). To calculate the z-score, z=x−μσ=68−644=44=1 This means that x=68 is one standard deviations (σ) above or to the right of the mean. This makes sense because the standard deviation is 4. So, one standard deviation would be (1)(4)=4, which is the distance between the mean (μ=64) and the value of x (68)

Suppose X∼N(5.5,2), and x=7.5. Find and interpret the z-score of the standardized normal random variable.

This means that x=7.5 is one standard deviation (1σ) above or to the right of the mean, μ=5.5. X is a normally distributed random variable with μ=5.5 (mean) and σ=2 (standard deviation). To calculate the z-score, z=x−μσ=7.5−5.52=1 This means that x=7.5 is one standard deviations (1σ) above or to the right of the mean. This makes sense because the standard deviation is 2. So, one standard deviation would be (2)(1)=2, which is the distance between the mean (μ=5.5) and the value of x (7.5).

Lexie scores an average of 527 points in a pinball game, and her points in a pinball game are normally distributed. Suppose Lexie scores 471 points in a game, and this value has a z-score of −2. What is the standard deviation? Do not include the units in your answer. For example, if you found that the standard deviation was 10 points, you would enter 10

We can work backwards using the z-score formula to find the standard deviation. The problem gives us the values for z, x and μ. So, let's substitute these numbers back into the formula: z−2−2−2σσ=x−μσ=471−527σ=−56σ=−56=28 We can think of this conceptually as well. We know that the z-score is −2, which tells us that x=471 is two standard deviations to the left of the mean, 527. So we can think of the distance between 471 and 527, which is 56. This distance is broken up into two standard deviations, so 562=28. So 28 is the standard deviation.