a resistencia de vigas de madeira utilizadas na construcao esta sendo estudada o fornecedor ates

To find the probability of a single beam supporting less than 1 ton, we need to standardize the

A resistencia de vigas de madeira utilizadas na construcao...

A resistência de vigas de madeira utilizadas na construção está sendo estudada. O fornecedor atesta que, em média, cada viga resiste a 3 toneladas com desvio padrão de aproximadamente 2 toneladas. Vinte dessas vigas serão sorteadas do fornecedor e supondo que o modelo Normal é adequado, pergunta-se: a. Qual a probabilidade de uma dessas vigas suportar menos do que 1 tonelada? b. Qual a probabilidade de as vinte vigas suportarem, em média, pelo menos 2,5 toneladas? c. Qual a probabilidade em (b), considerando agora 40 vigas e sem fazer a suposição de normalidade para os dados?

Transcript

00:01 Hi, i'm david and i'm here to help you answer your question.

00:03 Now let me bring up your question here.

00:06 In this question we're going to review about the normal distribution.

00:11 Let me remind you that if we have the x followed by the normal distribution with the mean and standard deviation.

00:19 And then we can have the x minus the mean divided by standard deviation.

00:25 It will equal to the standard normal.

00:27 And if we consider the x part minus the standard deviation, mean over standard division, we need to divide by square of n, in order to obtain the standard normal.

00:37 Now, if x does not follow by the normal, and then by central limit theorem, if n greater equal to the 30, then we still have xx minus m u over sigma over square root of n, and it's still followed by the normal.

00:55 And in this question here, we're given that the mean it would equal to the number.

01:01 To the 3 and the standard division sigma equal to the 2.

01:06 Now we have n equal to the 20 of the spins, so end equal to the 20.

01:12 Now we assume that x will follow by the normal.

01:16 So therefore in the question a, once you find the probability that one of these spins which is less than one tonne, so x will be smaller than 1.

01:26 To find this probability, i applied the formula for the first single 1x.

01:32 And turn into the z by writing the 1 minus the mean divided by standard deviation.

01:40 Now we get equal to the z smaller than the minus 1.

01:45 Now to find this probability, i will bring up the z table here.

01:50 So let me copy the z table.

01:52 I will show you how to use the z table here.

01:56 I will put the table on the right.

01:58 Now with the z smaller than minus 1, we see the z score equal to minus 1...