Question

**Question 1: In a large city, 46% of adults support the
local football team building a new stadium. If a poll is taken from
a random sample of 80 adults in the large city, which of the
following properly describes the sampling distribution of the
sample proportion of adults who support the stadium?**

A) Mean: 36.8, Sigma P-hat: 4.46, the distribution is approximately normal.

B) Mean: 36.8, Sigma P-hat: 4.46, shape of the distribution is unknown.

C) Mean: 0.46, Sigma P-hat: 0.056, the distribution is approximately normal.

D) Mean: 0.46, Sigma P-hat: 0.056, shape of the distribution is unknown.

E) Mean: 43.2, Sigma P-hat: 4.46, the distribution is binomial.

**Question 2: Which of the following statements is
true?**

A) A parameter is a number that describes some characteristic of a sample.

B) An unbiased estimator is any statistic that is taken from a sample chosen by random methods.

C) A sampling distribution is the distribution of a statistic calculated from all possible samples of the same size from the same population.

D) The variability of a population distribution will decrease as the sample size increases.

E) A normal approximation can always be used for the sampling distribution of P-hat as long as the sample size is greater than 30

**Question 3: The weight of a single bag checked by an
airplane passenger follows a distribution that is right skewed with
a mean of 38 pounds and a standard deviation of 6.2 pounds. If a
random sample of 96 bags is selected, what is the probability that
the average weight of the bags exceeds 40 pounds?**

A) 0.0008

B) 0.0011

C) 0.3735

D) 0.9992

E) It is not appropriate to use a normal distribution to calculate probability in this situation.

Answer #1

1)

np= | 36.8000 |

n(1-p)= | 43.2000 |

here population proportion= p= | 0.4600 | |

sample size =n= | 80 | |

std error of proportion=σ_{p}=√(p*(1-p)/n)= |
0.056 |

C) Mean: 0.46, Sigma P-hat: 0.056, the distribution is approximately normal.

2)

C) A sampling distribution is the distribution of a statistic calculated from all possible samples of the same size from the same population.

3)

for normal distribution z score =(X-μ)/σ | |

here mean= μ= | 38 |

std deviation =σ= | 6.2000 |

sample size =n= | 96 |

std error=σ_{x̅}=σ/√n= |
0.6328 |

probability
=P(X>40)=P(Z>(40-38)/0.633)=P(Z>3.16)=1-P(Z<3.16)=1-0.9992=0.00080 |

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