91影视

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

For a randomly selected female-male pair of mating scorpion flies, \({\rm{X = }}\)courting time.

Because a negative time makes no sense in this scenario, the shortest conceivable time is \({\rm{0}}\) minutes.

The duration might be as long as possible, implying that the distribution is probably skewed to the right (or positively skewed).

Because a normal distribution can take on negative values and is not skewed, \({\rm{X}}\)cannot be considered a normal distribution

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

\({\rm{n = 50}}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), the sample mean \({\rm{\bar x}}\)sampling distribution is approximately normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar x}}\)is nearly normal because the sample size is greater than \({\rm{30}}\).

The sample mean's sampling distribution has a mean of \({\rm{\mu }}\)and a standard deviation of \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

The \({\rm{z}}\)-value is the sample mean divided by the standard deviation of the population mean:

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{100 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg - 1}}{\rm{.29}}}\\{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{125 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg 0}}{\rm{.32}}}\end{array}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of \({\rm{z}}\)-scores).

\begin{aligned}P(100 拢 \bar X 拢125) &= P( - 1.29 < Z < 0.32) = P(Z < 0.32) - P(Z < - 1.29) \\& = 0.6255 - 0.0985 = 0.5270 = 52.70\% \end{aligned}

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\end{array}\)

\({\rm{n = 50}}\)

The total duration is \({\rm{150}}\)hours, or \({\rm{150 \times 60 = 9000}}\)minutes. The sample mean is calculated by dividing the total time by the sample size:

\({\rm{\bar x = }}\frac{{{\rm{9000}}}}{{{\rm{50}}}}{\rm{ = 180}}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), the sample mean \({\rm{\bar x}}\)sampling distribution is approximately normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar x}}\) is nearly normal because the sample size is greater than \({\rm{30}}\).

The sample mean's sampling distribution has a mean of \({\rm{\mu }}\)and a standard deviation of \(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\)

The \({\rm{z}}\)-value is the sample mean divided by the standard deviation of the population mean:

\({\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{{\rm{\sigma /}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{{\rm{180 - 120}}}}{{{\rm{110/}}\sqrt {{\rm{50}}} }}{\rm{\gg 3}}{\rm{.86}}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of \({\rm{z}}\)-scores).

\(P\left( {{a^{\circ } }{X_i} > 150} \right) = P(Z > 3.86) = 1 - P(Z < 3.86)\gg 1 - 1 = 0\)

Given:

\(\begin{array}{*{20}{c}}{{\rm{\mu = 120min}}}\\{{\rm{\sigma = 110min}}}\\{{\rm{n = 50}}}\\{{\rm{n = 15}}}\end{array}\)

The central limit theorem states that if the sample size is big (more than \({\rm{30}}\)), the sample mean \({\rm{\bar x}}\)sampling distribution is approximately normal.

The central limit theorem tells us that the sampling distribution of the sample mean \({\rm{\bar x}}\) is nearly normal because the sample size is greater than \({\rm{30}}\).

We can't utilise the central limit theorem because the sample size is smaller than 30. It is not possible to compute the requested probability since the distribution of \({\rm{X}}\)is not nearly normal (by part (a)) and we do not know the distribution of \({\rm{X}}\)