The number of points scored by each team in the NHL at the end of the season is

normally distribute with a mean of 89 and standard deviation of 11. Find P(x<85).

Answer:

The probability that [tex] \\ P(x<85)[/tex] is, approximately, 0.3594 or about 35.94% (or simply 36%).

Step-by-step explanation:

Firstly, we have to know that the random variable, in this case, is normally distributed. A normal distribution is completely determined by its two parameters,  namely, the population mean and the population standard deviation. For the case in question, we have a mean of [tex] \\ \mu = 89[/tex], and a standard deviation of [tex] \\ \sigma = 11[/tex].

To find the probability in question, we can use the standard normal distribution, a special case of a normal distribution with mean equals 0 and a standard deviation that equals 1.

All we have to do is "transform" the value of the raw score (x in this case) into its equivalent z-score. In other words, we first standardize the value x, and then we can find the corresponding probability.

With this value, we can consult the cumulative standard normal table, available in most Statistics textbooks or on the Internet. We can also make use of technology and find this probability using statistical packages, spreadsheets, and even calculators.

The corresponding z-score for a raw score is given by the formula:

[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]

And it tells us the distance of the raw score from the mean in standard deviations units. A positive value of the z-score indicates that the raw value is above the mean. Conversely, a negative value tells us that the raw score is below the mean.

With all this information, we are prepared to answer the question.

The corresponding z-score

According to formula [1], the z-score, or standardized value, is as follows:

[tex] \\ z = \frac{x - \mu}{\sigma}[/tex]

From the question, we already know that

x = 85

[tex] \\ \mu = 89[/tex]

[tex] \\ \sigma = 11[/tex]

Thus

[tex] \\ z = \frac{85 - 89}{11}[/tex]

[tex] \\ z = \frac{-4}{11}[/tex]

[tex] \\ z = -0.3636 \approx -0.36[/tex]

Consult the probability using the cumulative standard normal table

With this value for z = -0.36, we can consult the cumulative standard normal table. The entry for use it is this z-score with two decimals. This z-score tells us that the raw value of 85 is 0.36 standard deviations below from the population mean.

For z = -0.36, we can see, in the table, an initial entry of -0.3 at the first column of it. We then need to find, in the first line (or row) of the table, the corresponding 0.06 decimal value remaining. With this two values, we can determine that the cumulative probability is, approximately:

[tex] \\ P(x<85) = P(z<-0.36) = 0.3594[/tex]

Then, [tex] \\ P(x<85) = 0.3594[/tex] or, in words, the probability that [tex] \\ P(x<85)[/tex] is, approximately, 0.3594 or about 35.94% (or simply 36%).

Remember that this probability is approximated since we have to round the value of z to two decimal places (z = -0.36), and not (z = -0.3636), because of the restrictions to two decimals places for z of the standard normal table. A more precise result is [tex] \\ P(x<85) = 0.3581[/tex] using technology, shown in the graph below.

Notice that, in the case that the cumulative standard normal table does not present negative values for z, we can use the next property of the normal distributions, mainly because of the symmetry of this family of distributions.

[tex] \\ P(z<-a) = 1 - P(z<a) = P(z>a)[/tex]

For the case presented here, we have

[tex] \\ P(z<-0.36) = 1 - P(z<0.36) = P(z>0.36)[/tex]

[tex] \\ P(z<-0.36) = 1 - 0.6406 = P(z>0.36)[/tex]

[tex] \\ P(z<-0.36) = 0.3594 = P(z>0.36)[/tex]

Which is the same probability obtained in the previous step.

The graph below shows the shaded area for the probability of  [tex] \\ P(x<85)[/tex] finally obtained.  

Final answer:

To find P(x<85), calculate the z-score and use the standard normal distribution to find the cumulative probability. Susan's z-score for her final exam is 2, meaning her performance was significantly above the average. The central limit theorem helps to analyze the mean of large sample sizes, while binomial distributions apply to scenarios such as analyzing a basketball player's field goal completion rate.

Explanation:

Finding Probability in a Normal Distribution

To find the probability P(x<85) when the number of points scored by each team in the NHL at the end of the season is normally distributed with a mean (μ) of 89 and a standard deviation (σ) of 11, we use the standard normal distribution. First, we calculate the z-score for x=85, which is the value for which we want to find the cumulative probability. The z-score is calculated using the formula z = (x - μ) / σ. Substituting the given values, we get z = (85 - 89) / 11 = -0.3636. Then we look up this z-score in the standard normal distribution table or use a calculator or software to find the cumulative probability associated with this z-score.

To express the number 13.7 in terms of the mean and standard deviation of the given data, you would use the formula for the z-score again.

If Susan's biology class has a mean final exam score of 85 and a standard deviation of 5, and Susan scored a 95 on her final exam, her z-score would be z = (95 - 85) / 5 = 2. This means Susan's score is 2 standard deviations above the mean, indicating she performed significantly better than the average student.

When dealing with sample sizes larger than one, the central limit theorem tells us that the sampling distribution of the sample mean will tend to be normal regardless of the shape of the population distribution, especially as the sample size increases (typically n > 30 is considered large enough). This is reflected in the examples involving the estimation of mean final exam scores and calculating probabilities with a sample size of 55.

The probability distribution question for the basketball player's shots would involve the binomial distribution, where the probability of success is given by the player's field goal completion rate. The mean and standard deviation for this binomial distribution can be calculated using the formulas for a binomial distribution.

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