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Chapter 4: Problem 11

Assume that 50 births are randomly selected. Use subjective judgment to describe the given number of girls as (a) significantly low, (b) significantly high, or (c) neither significantly low nor significantly high. 23 girls.

Short Answer

Expert verified
23 girls is neither significantly low nor significantly high.

Step by step solution

01

- Define the Parameters

In this problem, each birth can be seen as a Bernoulli trial with two possible outcomes: girl or boy. Let's define our parameters: the number of trials (births) is 50, and the probability of having a girl in any single birth is 0.5 (assuming an equal probability of having a girl or a boy).
02

- Calculate the Mean and Standard Deviation

For a binomial distribution, the mean (expected number of girls) is given by \(\text{mean} = n \times p\) and the standard deviation is given by \(\text{standard deviation} = \sqrt{n \times p \times (1-p)}\). Substituting the values, we get \(\text{mean} = 50 \times 0.5 = 25\) and \(\text{standard deviation} = \sqrt{50 \times 0.5 \times 0.5} = 3.54\).
03

- Determine Significance Using Z-scores

Next, calculate the z-score for 23 girls to determine if this number is significantly low, high, or neither. The z-score can be calculated using the formula \(\text{z} = \frac{x - \text{mean}}{\text{standard deviation}}\). Substituting the values, we get \(\text{z} = \frac{23 - 25}{3.54} = -0.57\).
04

- Interpret the Z-score

A z-score between -2 and 2 is considered to be within the range of typical values, meaning it is neither significantly low nor significantly high. Since -0.57 falls within this range, the number of 23 girls is neither significantly low nor significantly high.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bernoulli trial
A Bernoulli trial is a random experiment where there are only two possible outcomes: success or failure. For example, in each birth, we can either have a girl (success) or a boy (failure). Each birth here is considered independent of others, meaning the gender of one child doesn't affect the gender of another.
Bernoulli trials are fundamental in binomial distributions, where events can be repeated multiple times, and each event retains the same probability of success.
mean and standard deviation
The mean and standard deviation are essential in understanding distributions. For a binomial distribution, the mean (also called the expected value) is found using \text{mean} = n \times p\, where 'n' is the number of trials, and 'p' is the probability of success in each trial.
In the given exercise, the number of births (trials) is 50, and the probability of having a girl (success) in each birth is 0.5. Hence, the mean is \text{mean} = 50 \times 0.5 = 25\.
The standard deviation measures how much variation there is from the mean. It's calculated using the formula:\[ \text{standard deviation} = \sqrt{n \times p \times (1-p)}\]. For our problem, \text{standard deviation} = \sqrt{50 \times 0.5 \times 0.5} = 3.54\.
z-score interpretation
The z-score quantifies how far an individual data point is from the mean, measured in terms of the standard deviation. It's calculated as:\( \text{z} = \frac{x - \text{mean}}{\text{standard deviation}}\), where 'x' is your data point.
For the 23 girls in our example, we calculate the z-score as:\(\text{z} = \frac{23 - 25}{3.54} = -0.57\). A z-score between -2 and 2 is generally considered typical, meaning the number of girls isn't significantly different from what we would expect.
statistical significance
Statistical significance helps us determine whether an observed effect is due to chance. In this context, we use the z-score to assess whether 23 girls out of 50 births is unusual. Typically, if the z-score is beyond -2 or 2, we consider it statistically significant. In our case, the z-score of -0.57 lies within the range of -2 to 2, suggesting that getting 23 girls is neither significantly low nor high; hence, it is not statistically unusual.
Understanding statistical significance is crucial for making inferences in any scientific study or practical decision-making.

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Most popular questions from this chapter

Express all probabilities as fractions. Current rules for telephone area codes allow the use of digits \(2-9\) for the first digit, and \(0-9\) for the second and third digits. How many different area codes are possible with these rules? That same rule applies to the exchange numbers, which are the three digits immediately preceding the last four digits of a phone number. Given both of those rules, how many 10 -digit phone numbers are possible? Given that these rules apply to the United States and Canada and a few islands, are there enough possible phone numbers? (Assume that the combined population is about \(400,000,000 .\) )

Involve redundancy. It is generally recognized that it is wise to back up computer data. Assume that there is a \(3 \%\) rate of disk drive failure in a year (based on data from various sources, including lifehacker.com). a. If you store all of your computer data on a single hard disk drive, what is the probability that the drive will fail during a year? b. If all of your computer data are stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that both drives will fail during a year? c. If copies of all of your computer data are stored on three independent hard disk drives, what is the probability that all three will fail during a year?

A study of the effect of seatbelt use in head-on passenger car collisions found that drivers using a seatbelt had a \(64.1 \%\) survival rate, while drivers not using a seatbelt had a \(41.5 \%\) survival rate. If seatbelts have no effect on survival rate, there is less than a \(0.0001\) chance of getting these results (based on data from "Mortality Reduction with Air Bag and Seat Belt Use in Head-on Passenger Car Collisions," by Crandall, Olson, Sklar, American Journal of Epidemiology, Vol. 153, No. 3 ). What do you conclude?

Find the probability and answer the questions. When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the expected value of \(3 / 4\), as Mendel claimed?

Express all probabilities as fractions. Mendel conducted some his famous experiments with peas that were either smooth yellow plants or wrinkly green plants. If four peas are randomly selected from a batch consisting of four smooth yellow plants and four wrinkly green plants, find the probability that the four selected peas are of the same type.
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