Calculating Probabilities for Z-Scores Using the Standard Normal Distribution

Calculating Probabilities for Z-Scores Using the Standard Normal Distribution

Question 1

To find the probabilities for a given z-score, one would typically use a z-table or software that provides the cumulative distribution function (CDF) for the standard normal distribution. For the given questions, here are the probabilities:

1. The probability between (z = -0.52) and (z = 2.33) is:
   = P(-0.52 < z < 2.33)

= 0.6886

1. The probability for (z > 1.65) is:

= P(z > 1.65)

= 0.0495

1. The probability for (z < 1.85) is:

= P(z < 1.85)

= 0.9678

1. The probability for (z < 0) is:

= P(z < 0)

= 0.5000

This is because the mean of the standard normal distribution is 0.

1. The probability between (z = 0) and (z = 1 \) is:

= P(0 < z < 1)

= 1.0000

Question 2

1. The distribution of the means computed from samples of a specific size selected from a population is called the sampling distribution of the mean. It represents all possible values of the sample mean that could occur if many samples of that size were drawn from the population.
2. Most sample means differ somewhat from the population mean due to random sampling fluctuations or chance variations in the samples. This difference between a sample mean and the population mean is called the sampling error (Mishra et al., 2019). It arises because any given sample may not perfectly represent the entire population’s characteristics.
3. The mean of the sampling distribution of the mean is equal to the population mean itself. In other words, if you took the mean of all possible sample means, it would equal the true population mean from which the samples were drawn.
4. The standard deviation of the sampling distribution of the mean is called the standard error of the mean. It measures how much variability there is among the sample means around the true population mean (McGrath et al., 2020). The formula is: Standard Error = (Population Standard Deviation) / √(Sample Size). As the sample size increases, the standard error decreases.
5. The central limit theorem states that the sampling distribution of the mean approximates a normal distribution as the sample size increases, regardless of the original population distribution shape. This means the distribution of sample means becomes increasingly bell-shaped and symmetric around the population mean with larger samples, allowing statistical inferences.

Question 3

We are given the distribution of heights as follows:

 

The probability that the height is less than 34 inches:

Converting it to a standard normal variable, we have:

Using the standard normal tables, we have:

Therefore, the percentage of the three-year-old females that have a height less than 34 inches is 11.31%.

Question 4

Given n = 300

= 16.92

Since  > 10, we can use normal approximation to binomial

ơ =

=

Now,

= P (

Question 5

1. For the scenario of tossing a coin 16 times and calculating the proportion of heads, the shape of the histogram would be expected to be binomial or approximately bell-shaped. This is because each toss is a Bernoulli trial with two possible outcomes (heads or tails), and the proportion of heads is a binomial random variable with n=16 trials (Branson & Marie-Abèle Bind, 2017).
2. The histogram would be expected to be centered around 0.5, which is the theoretical probability of getting heads on a fair coin toss. Since the coin is fair and the tosses are independent, the expected value of the proportion of heads is 0.5
3. The variability among the proportions of heads can be calculated using the formula for the standard deviation of a binomial distribution.

Where, p = 0.5 and q = 1-p = 0.5

With these values,

Hence, variability is 0.125

With p=0.5 (probability of heads) and n=16 tosses, the standard deviation is approximately 0.125. This means that most of the observed proportions of heads will fall within a range of about 0.5 ± 0.25 (0.25 to 0.75).

1. A Normal model should not be used here because the proportion of heads is a discrete random variable that follows a binomial distribution, not a continuous normal distribution. The binomial distribution is more appropriate for modeling the number of successes (heads) in a fixed number of independent Bernoulli trials. The normal approximation to the binomial distribution may be valid for large values of n, but for a small number of trials like 16, the binomial distribution should be used directly.

Question 6

Let x be the no. of eggs.

Thus, by the central limit theorem,

)

Therefore, the probability that the cholesterol content will be greater than 230 milligrams. is 0.0217.

References

Branson, Z., & Marie-Abèle Bind. (2017). Randomization-based Inference for Bernoulli-Trial Experiments and Implications for Observational Studies. ArXiv (Cornell University). https://doi.org/10.48550/arxiv.1707.04136

McGrath, S., Zhao, X., Steele, R., Thombs, B. D., Benedetti, A., Levis, B., Riehm, K. E., Saadat, N., Levis, A. W., Azar, M., Rice, D. B., Sun, Y., Krishnan, A., He, C., Wu, Y., Bhandari, P. M., Neupane, D., Imran, M., Boruff, J., & Cuijpers, P. (2020). Estimating the sample mean and standard deviation from commonly reported quantiles in meta-analysis. Statistical Methods in Medical Research, 29(9), 2520–2537. https://doi.org/10.1177/0962280219889080

Mishra, P., Pandey, C. M., Singh, U., Sahu, C., Keshri, A., & Gupta, A. (2019). Descriptive statistics and normality tests for statistical data. Annals of Cardiac Anaesthesia, 22(1), 67–72. https://doi.org/10.4103%_157_18

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Week 3 Assignment

The following assignment will allow you to master the material on Normal Distributions & Sampling Distributions.

1. It is very important to be able to find probabilities for a given z score. Using software or a Standard Normal Distribution table, determine the following probability for the area (20 points)

a. between z = -0.52 and z = 2.33
b. z > 1.65
c. z < 1.85
d. z < 0
e. between z = 0 and z = 1

2. Answer the following: (20 points)

a. If samples of a specific size are selected from a population and the means are computed, what is the distribution of the means called?
b. Why do most of the sample means differ somewhat from the population mean? What is this difference called?
c. What is the mean of the sample means?
d. What is the standard deviation of the sample means called? What is the formula for this standard deviation?
e. What does the central limit theorem say about the shape of the distribution of sample means?

3. A pediatrician obtains the heights of her 200 three-year-old female patients. The heights are normally distributed with a mean of 37.8 and a standard deviation of 3.14. What percent of the three-year-old females have a height of less than 34 inches? (20 points)

4. The local news reported that 6% of U.S. drivers text on their cell phones while driving. If 300 drivers are selected at random, find the probability that exactly 25 say they text while driving. Be sure and use the continuity correction for the normal approximation to the binomial distribution. (15 points)

5. Suppose you are asked to toss a coin 16 times and calculate the proportion of the tosses that were heads. (15 points)

a. What shape would you expect this histogram to be and why?
b. Where you do expect the histogram to be centered?
c. How much variability would you expect among these proportions?
d. Explain why a Normal model should not be used here.

6. The average cholesterol content of a certain brand of eggs is 210 milligrams, and the standard deviation is 14 milligrams. Assume the variable is normally distributed. If two eggs are selected, find the probability that the cholesterol content will be greater than 230 milligrams. (10 points)