Birth weights Researchers in Norway analyzed data on the birth weights of 400,000 newborns over a 6-year period. The distribution of birth weights is approximately Normal with a mean of 3668 grams and a standard deviation of 511 grams.

a. Sketch a graph that displays the distribution of birth weights for this population.

b. Sketch a possible graph of the distribution of birth weights for an SRS of size $5 . Calculate the range for this sample.

In this population, the range (Maximum - Minimum) of birth weights is 3417 grams. We technology to take 500 SRSs of size n=5n=5and calculate the range (Maximum Minimum) for each sample. The dotplot shows the results.

Short Answer

Expert verified

(a)The middle 99.7%of a normal distribution lies with three standard deviation from the mean of the distribution.

(b)A possible sample of size 5 is then 2400,3000,3600,3700,4500Range=max-min=4500-2400=2100

(c) The dots is the dot plot represent the sample range of simple random samples of size n=5. The dot at 2800 then represents a single simple random sample size n=5that has a sample ranges of 2800 grams.

(d) it is observed that all sample ranges are smaller than the population range of 3066 grams. This then implies that the centre of the sampling distribution of the sample ranges is not the population range and thus the sample range is not an unbiased estimator of the population range.

Step by step solution

01

Part (a) Step 1: Given Information

Given,

μ=3668σ=511

02

Part (a) Step 2: Simplification

With one standard deviation from the distribution's mean, the centre 68 percent of a normal distribution lies.

The middle 95 percentile of a normal distribution is two standard deviations off the mean.

With three standard deviations from the distribution's mean, the middle 99.7% of a normal distribution is found.

Identifying values that are 1,2, and 3 standard deviations apart from the mean

μ-3σ=3668-3(511)=2135μ-2σ=3668-2(511)=2646μ-σ=3668-511=3157μ+σ=3668+511=4179μ+2σ=3668+2(511)=4690μ+3σ=3668+3(511)=5201

03

Part (b) Step 1: Given Information

Normal distribution

μ=3668σ=511

04

Part (b) Step 2: Simplification

A basic random sample of size 5 has 5 data values that are broadly centred around a, with all data values falling between and. μ-3σ=3668-3(511)=2135and μ+3σ=3668+3(511)=5201.

The size of a probable sample of size 5 is then calculated.2400,3000,3600,3700,4500.

The Range =max-min=4500-2400=2100

Place a dot above the associating number on the number line for each given data value.

05

Part (c) Step 1: Given Information

Given , The graph is

06

Part (c) Step 2: Simplification

The dots on the dot plot show the sample range of simple random samples with a size of n=5. A single simple random sample size n=5with a sample range of 2800 grammes is represented by the dot at 2800.

07

Part (d) Step 1: Given Information

Given,

μ=3668σ=511

08

Part (d) Step 2: Simplification

A normal distribution's middle 68 percent are one standard deviation from the distribution's mean. The middle 95 percentile of a normal distribution is two standard deviations off the mean. With three standard deviations from the distribution's mean, the middle 99.7% of a normal distribution is found.

μ-3σ=3668-3(511)=2135μ+3σ=3668+3(511)=5201

As a result, almost all data values fall between 5201 and 2135.

The difference between the maximum and minimum called the range.

Range=max-min=5201-2135=3066

This means that the population range is approximately 3066 grammes.

If the centre of the estimator's sample distribution equals the corresponding population parameter, the estimator is said to be unbiased.

All sample ranges are smaller than the population range of 3066 grammes, as seen in the dot plot. This suggests that the sample range's centre of sampling distribution is not the population range, and hence the sample range is not an unbiased estimator of the population range.

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Most popular questions from this chapter

Which of the following statements about the sampling distribution of the sample mean is incorrect?

a. The standard deviation of the sampling distribution will decrease as the sample size increases.

b. The standard deviation of the sampling distribution measures how far the sample mean typically varies from the population mean.

c. The sample mean is an unbiased estimator of the population mean.

d. The sampling distribution shows how the sample mean is distributed around the population mean.

e. The sampling distribution shows how the sample is distributed around the sample mean.

The number of flaws per square yard in a type of carpet material varies with mean 1.6flaws per square yard and standard deviation flaws per square yard.

a. Without doing any calculations, explain which event is more likely:

  • randomly selecting a1square yard of material and finding 2 or more flaws
  • randomly selecting 50square yards of material and finding an average of 2or more flaws

b. Explain why you cannot use a Normal distribution to calculate the probability of the first event in part (a).

c. Calculate the probability of the second event in part (a).

At a traveling carnival, a popular game is called the “Cash Grab.” In this game, participants step into a sealed booth, a powerful fan turns on, and dollar bills are dropped from the ceiling. A customer has 30 seconds to grab as much cash as possible while the dollar bills swirl around. Over time, the operators of the game have determined that the mean amount grabbed is \(13 with a standard deviation of \)9. They charge \(15 to play the game and expect to have 40 customers at their next carnival.

a. What is the probability that an SRS of 40 customers grab an average of \)15 or more?

b. How much should the operators charge if they want to be 95% certain that the mean amount grabbed by an SRS of 40 customers is less than what they charge to play the game?

The number of unbroken charcoal briquets in a 20-pound bag filled at the factory follows a Normal distribution with a mean of 450 briquets and a standard deviation of 20 briquets. The company expects that a certain number of the bags will be underfilled, so the company will replace for free the 5%of bags that have too few briquets. What is

the minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free?

a. 404

b. 411

c. 418

d. 425

e. 448

The central limit theorem is important in statistics because it allows us to use a Normal distribution to find probabilities involving the sample mean if the

a. sample size is reasonably large (for any population).

b. population is Normally distributed (for any sample size).

c. population is Normally distributed and the sample size is reasonably large.

d. population is Normally distributed and the population standard deviation is known (for any sample size).

e. population size is reasonably large (whether the population distribution is known or not).

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