91影视

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

a)

Given: \(\text{X}\)has a normal distribution with

\(\text{ }\!\!\mu\!\!\text{ =8}\text{.46 }\!\!~\!\!\text{ min }\!\!\sigma\!\!\text{ =0}\text{.913 }\!\!~\!\!\text{ min}\)

Complement rule:

\(\text{P(notA)=1-P(A)}\)

The standardized score is calculated by dividing the value \(\text{x}\)by the mean and then by the standard deviation.

\(\begin{aligned}& \text{z=}\frac{\text{x- }\!\!\mu\!\!\text{ }}{\text{ }\!\!\sigma\!\!\text{ }} \\ & \text{=}\frac{\text{10-8}\text{.46}}{\text{0}\text{.913}} \\& \text{ }\!\!\gg\!\!\text{ 1}\text{.69} \end{aligned}\)

Determine the equivalent probability in the appendix's normal probability table (which contains the probabilities to the left of z-score), which is supplied in the row with \(\text{1}\text{.6}\)and the column with. The complement rule is also used.

\(\begin{aligned} &\text{P(X }\!\!{}^\text{3}\!\!\text{ 10)=P(Z1}\text{.69)} \\ & \text{=1-P(Z1}\text{.69)=1-0}\text{.9545} \\ & \text{=0}\text{.0455=4}\text{.55 }\!\!%\!\!\text{ } \\ \end{aligned}\)

Therefore, the probability is \(P(X\ge 10)=4.55%\)

2)

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{aligned} z &= \frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\ &= \frac{{{\rm{15 - 8}}{\rm{.46}}}}{{{\rm{0}}{\rm{.913}}}}\\{\rm{\gg 7}}{\rm{.16}}\end{aligned}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probabilities to the left of z-score). We may presume that the probability to the left is l since the z-score of \({\rm{7}}{\rm{.16}}\)is not stated in the table and is bigger than all other numbers. The complement rule is also used.

\(\begin{aligned}P(X > 15) &= P(Z > 7 {\rm{.16)}}\\ &= 1 - P(Z < 7{\rm{.16)}}\\ gg 1 - 1 &= 0 \\ &= 0\% \end{aligned}\)

Therefore, the probability is \({\rm{P(X > 15) = 0 = 0\% }}\).

3)

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{8 - 8}}{\rm{.46}}}}{{{\rm{0}}{\rm{.913}}}}\\{\rm{\gg - 0}}{\rm{.50}}\\{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{10 - 8}}{\rm{.46}}}}{{{\rm{0}}{\rm{.913}}}}\\{\rm{\gg 1}}{\rm{.69}}\end{array}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probabilities to the left of z-score).

\(\begin{aligned}P(8 < X < 10) &= P( - 0{\rm{.50 < Z < 1}}{\rm{.69)}}\\ &= P(Z < 1{\rm{.69) - P(Z < - 0{\rm{.50)n}}\\ &= 0 {\rm{.9545 - 0}}{\rm{.3085}}\\ &= 0 {\rm{.6460}}\\ &= 64 {\rm{.60\% }}\end{aligned}\)

4)

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{8}}{\rm{.46 \pm c - 8}}{\rm{.46}}}}{{{\rm{0}}{\rm{.913}}}}\\{\rm{\gg }}\frac{{{\rm{ \pm c}}}}{{{\rm{0}}{\rm{.913}}}}\end{array}\)

If there is \({\rm{98\% }}\)of all possible times between \({\rm{8}}{\rm{.46 - c}}\)and \({\rm{8}}{\rm{.46 + c}}\) since the normal distribution is symmetric about its mean\({\rm{\mu = 8}}{\rm{.46}}\), then \({\rm{1\% }}\)of the possible times is below \({\rm{8}}{\rm{.46 - c}}\)and \({\rm{1\% + 98\% = 99\% }}\)of the possible times is below \({\rm{8}}{\rm{.46 + c}}{\rm{.}}\)

Determine the z-score in the normal probability table in the appendix (which contains the probabilities to the left of z-score) that corresponds with a probability of\({\rm{1\backslash \% or 0}}{\rm{.01}}\).

\({\rm{z = - 2}}{\rm{.33}}\)

Determine the \({\rm{z}}\) -score in the normal probability table in the appendix (which contains the probabilities to the left of \({\rm{z}}\)-score) that corresponds with a probability of\({\rm{99\% or 0}}{\rm{.99}}\),

\({\rm{z = 2}}{\rm{.33}}\)

These \({\rm{z}}\)-scores have to be equal to the expression found for the \({\rm{z}}\)-scores:

\(\frac{{{\rm{ \pm c}}}}{{{\rm{0}}{\rm{.913}}}}{\rm{ = \pm 2}}{\rm{.33}}\)

Multiply each side by\({\rm{0}}{\rm{.913}}\):

\(\begin{array}{c}{\rm{c = 2}}{\rm{.33 \times 0}}{\rm{.913}}\\{\rm{ = 2}}{\rm{.12729}}\end{array}\)

The value c is \({\rm{2}}{\rm{.12729}}\).

5)

The standardized score is calculated by dividing the value \({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{10 - 8}}{\rm{.46}}}}{{{\rm{0}}{\rm{.913}}}}\\{\rm{\gg 1}}{\rm{.69}}\end{array}\)

Use the normal probability table in the appendix (which contains the probabilities to the left of z-score) to get the relevant probability, which is given in the row with \({\rm{1}}{\rm{.6}}\)and the column with\({\rm{.09}}\).

\({\rm{P(X < 10) = P(Z < 1}}{\rm{.69) = 0}}{\rm{.9545}}\)

We may apply the multiplication formula for independent events if the four haul times are independent:

\(\begin{array}{c}{\rm{P( Four haul times < 10) = P(X < 10) \times P(X < 10) \times P(X < 10) \times P(X < 10)}}\\{\rm{ = (P(X < 10)}}{{\rm{)}}^{\rm{4}}}\\{\rm{ = 0}}{\rm{.954}}{{\rm{5}}^{\rm{4}}}\\{\rm{\gg 0}}{\rm{.830049}}\end{array}\)

Use the complement rule:

\(\begin{array}{c}{\rm{P( At least one time > 10) = 1 - P( Four haul times < 10) = 1 - 0}}{\rm{.830049}}\\{\rm{ = 0}}{\rm{.169951}}\\{\rm{ = 16}}{\rm{.9951\% }}\end{array}\)