2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection

3. A drinks machine dispenses coffee into cups. A sign on the machine indicates that each cup contains 50 ml of coffee. The machine actually dispenses a mean amount of 55 ml per cup and \(10 \%\) of the cups contain less than the amount stated on the sign. Assuming that the amount of coffee dispensed into each cup is normally distributed find

1. the standard deviation of the amount of coffee dispensed per cup in ml ,
2. the percentage of cups that contain more than 61 ml . Following complaints, the owners of the machine make adjustments. Only \(2.5 \%\) of cups now contain less than 50 ml . The standard deviation of the amount dispensed is reduced to 3 ml . Assuming that the amount of coffee dispensed is still normally distributed,
3. find the new mean amount of coffee per cup.
   (4)

7. The random variable \(X\) is normally distributed with mean 79 and variance 144 . Find

1. \(\mathrm { P } ( X < 70 )\),
2. \(\mathrm { P } ( 64 < X < 96 )\). It is known that \(\mathrm { P } ( 79 - a \leq X \leq 79 + b ) = 0.6463\). This information is shown in the figure below. \includegraphics[max width=\textwidth, alt={}, center]{df898ff4-c3ef-400c-b4f7-f4df3757941d-6_581_983_818_590} Given that \(\mathrm { P } ( X \geq 79 + b ) = 2 \mathrm { P } ( X \leq 79 - a )\),
3. show that the area of the shaded region is 0.1179 .
4. Find the value of \(b\).

7. The heights of a group of athletes are modelled by a normal distribution with mean 180 cm and a standard deviation 5.2 cm . The weights of this group of athletes are modelled by a normal distribution with mean 85 kg and standard deviation 7.1 kg . Find the probability that a randomly chosen athlete

1. is taller than 188 cm ,
2. weighs less than 97 kg .
   (2)
3. Assuming that for these athletes height and weight are independent, find the probability that a randomly chosen athlete is taller than 188 cm and weighs more than 97 kg .
4. Comment on the assumption that height and weight are independent.

5. A random variable \(X\) has a normal distribution.

1. Describe two features of the distribution of \(X\). A company produces electronic components which have life spans that are normally distributed. Only \(1 \%\) of the components have a life span less than 3500 hours and \(2.5 \%\) have a life span greater than 5500 hours.
2. Determine the mean and standard deviation of the life spans of the components. The company gives warranty of 4000 hours on the components.
3. Find the proportion of components that the company can expect to replace under the warranty.

5. A health club lets members use, on each visit, its facilities for as long as they wish. The club's records suggest that the length of a visit can be modelled by a normal distribution with mean 90 minutes. Only \(20 \%\) of members stay for more than 125 minutes.

1. Find the standard deviation of the normal distribution.
2. Find the probability that a visit lasts less than 25 minutes. The club introduce a closing time of 10:00 pm. Tara arrives at the club at 8:00 pm.
3. Explain whether or not this normal distribution is still a suitable model for the length of her visit.

5. From experience a high-jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts. Assuming that the heights the high-jumper can reach follow a Normal distribution,

1. draw a sketch to illustrate the above information,
2. find, to 3 decimal places, the mean and the standard deviation of the heights the high-jumper can reach,
3. calculate the probability that he can jump at least 1.74 m .

Each day, for many years, the maximum temperature in degrees Celsius at a particular location is recorded. The maximum temperatures for days in October can be modelled by a Normal distribution. The appropriate Normal curve is shown in Fig. 6. \includegraphics{figure_6}

1. 1. Use the model to write down the mean of the maximum temperatures. [1]
   2. Explain why the curve indicates that the standard deviation is approximately 3 degrees Celsius. [1]

Temperatures can be converted from Celsius to Fahrenheit using the formula \(F = 1.8C + 32\), where \(F\) is the temperature in degrees Fahrenheit and \(C\) is the temperature in degrees Celsius.

1. For maximum temperature in October in degrees Fahrenheit, estimate
   [2]