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A conceivable result of an experiment or trial is referred to as an outcome in probability theory. Each experiment's probable results are distinct, and they are mutually exclusive (only one outcome will occur on each trial of the experiment).

Let the contents of jars be represented by a RV X then its mean and standard deviation are given as:

\(\begin{array}{l}\mu = 137.2{\rm{oz}}\\\sigma = 1.6{\rm{oz}}\end{array}\)

(a) The stated contents are 135 oz. Hence the probability that a single jar contains more than the stated contents is denoted as\(P(135 < X)\). Standardizing gives:

\(135 < X\)if and only if

\(\frac{{135 - 137.2}}{{1.6}} < \frac{{X - 137.2}}{{1.6}}\frac{{ - 2.2}}{{1.6}} < \frac{{X - 137.2}}{{1.6}} - 1.375 < Z\)

Thus

\(P(135 < X) = P( - 1.375 < Z)\)

Here $Z$ is a standard normal distribution rv with cdf\(\phi (z)\). Hence

\(P(135 < X) = P( - 1.375 < Z) = 1 - \phi ( - 1.375)\)

To get\(\phi ( - 1.375)\), we use Appendix Table A.3 and interpolation. Use z-values of\( - 1.37\)and\( - 1.38\). Since\( - 1.375\)lies in between these two points hence :

\(\begin{aligned}\phi ( - 1.375) &= \frac{{\phi ( - 1.37) + \phi ( - 1.38)}}{2}\\ &= \frac{{0.0853 + 0.0838}}{2}\phi ( - 1.375)\\ &= 0.0846P(135 < X)\\ &= 1 - 0.0846 = 0.9154\end{aligned}\)

The probability that a single jar contains more than the stated contents is \(0.9154\)

(b) Let N denote the number of jars among ten randomly selected jars that contain more than the stated contents. Then the probability that at least eight contain more than the stated contents can be represented as\(P(N \ge 8)\)

The probability that a single jar contains more than the stated contents is \(0.9154\)(from last part). Then the probability that a single jar contains less than the stated contents is\((1 - 0.9154 = 0.0846)\)

\(P(N \ge 8) = P(N = 8) + P(N = 9) + P(N = 10)\)

\(= C_8^{10} \cdot {(0.9154)^8} \cdot {(0.0846)^2} + C_9^{10} \cdot {(0.9154)^9} \cdot (0.0846) + C_{10}^{10} \cdot {(0.9154)^{10}}\)

\(P(N \ge 8) = 0.9538\)

(c) If\(95\% \)of all jars contain more than the stated contents, then it means that the probability that a single jar contains more than the stated contents is\(0.95\).

\(P(135 < X) = 0.95\)

Standardizing gives:

\(135 < X\)

if and only if

\(135 - 137.2\sigma < \frac{{X - 137.2}}{\sigma }\frac{{ - 2.2}}{\sigma } < \frac{{X - 137.2}}{\sigma }{Z_o} < Z\)

Thus

\(P(135 < X) = P\left( {{Z_o} < Z} \right)\)

Here Z is a standrad normal distribution rv with\({\mathop{\rm cdf}\nolimits} \phi (z)\). Hence

\(P(135 < X) = P\left( {{Z_o} < Z} \right) = 1 - \phi \left( {{Z_o}} \right)\)

But this probability is given equal to\(0.95\), hence

\(\begin{aligned}0.95 &= 1 - \phi \left( {{Z_o}} \right) \hfill \\\phi \left( {{Z_o}} \right) &= 0.05 \hfill \\\end{aligned} \)

Now using Appendix Table A.3. We note that z-value corresponding to probability of\(0.05\)is approximately equal to\( - 1.645\)

\(\begin{aligned}{Z_o} &= - 1.645\frac{{ - 2.2}}{\sigma }\\ &= - 1.645\sigma \\ &= \frac{{2.2}}{{1.645}}\sigma \\ &= 1.3374\end{aligned}\)