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KatRina [158]
3 years ago
7

Which of the following is NOT a conclusion of the Central Limit​ Theorem? Choose the correct answer below. A. The mean of all sa

mple means is the population mean mu. B. The standard deviation of all sample means is the population standard deviation divided by the square root of the sample size. C. The distribution of the sample data will approach a normal distribution as the sample size increases. D. The distribution of the sample means x overbar ​will, as the sample size​ increases, approach a normal distribution.
Mathematics
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

The distribution of the sample data will approach a normal distribution as the sample size increases.

Step-by-step explanation:

Central limit​ theorem states that the mean of all samples from the same population will be almost equal to the mean of the population, if the large sample size from a population, is given with a finite level of variance.

So, here Option C is not correct conclusion of central limit theorem -The distribution of the sample data will approach a normal distribution as the sample size increases.

We can say that the average of sample mean tends to be normal but not the sample data.

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The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

7 0
3 years ago
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