1. Family income is believed to be normally distributed with a mean of shs 250000 and a standard deviation of 100000. If the poverty level is 100000, what percentage of the population lives in poverty? A new tax law is expected to benefit middle-income families, those with incomes between shs 200000 and shs 300000. What percentage of the population will benefit from the law?

2. A thermostat set to switch at 20°C operates at a range of temperatures having a mean of 20.4°C and a standard deviation of 1.3°C. Determine the probability of its opening at temperatures between 19.5°C and 20.5°C.

To solve the problems given, we will use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetrical around its mean.

Percentage of Population Living in Poverty:
Family income is normally distributed with a mean ([tex]\mu[/tex]) of 250,000 shillings and a standard deviation ([tex]\sigma[/tex]) of 100,000 shillings. We want to find the percentage of the population with incomes below the poverty level of 100,000 shillings.
First, we calculate the z-score, which is the number of standard deviations away from the mean:
[

z = \frac{X - \mu}{\sigma} = \frac{100,000 - 250,000}{100,000} = \frac{-150,000}{100,000} = -1.5
]

A z-score of -1.5 corresponds to a percentile in the standard normal distribution table. Looking this up in a z-table, we find that a z-score of -1.5 corresponds to approximately 6.68%. Thus, about 6.68% of the population lives in poverty.

Percentage of the Population Benefiting from the New Tax Law:
The tax law benefits those with incomes between 200,000 and 300,000 shillings.
We calculate the z-scores for these income levels:
For 200,000:
[

z_{200} = \frac{200,000 - 250,000}{100,000} = \frac{-50,000}{100,000} = -0.5
]

For 300,000:
[tex]z_{300} = \frac{300,000 - 250,000}{100,000} = \frac{50,000}{100,000} = 0.5[/tex]

A z-score of -0.5 corresponds to approximately 30.85% and a z-score of 0.5 corresponds to approximately 69.15%.

The percentage of the population with incomes between these two z-scores is:

(
69.15% - 30.85% = 38.3%
)

Therefore, 38.3% of the population is expected to benefit from the new tax law.

Probability of Thermostat Opening at 19.5°C to 20.5°C:
The thermostat operates with a mean temperature of 20.4°C and a standard deviation of 1.3°C.
We calculate the z-scores for the two temperatures:
For 19.5°C:
[

z_{19.5} = \frac{19.5 - 20.4}{1.3} = \frac{-0.9}{1.3} \approx -0.69
]

For 20.5°C:
[tex]z_{20.5} = \frac{20.5 - 20.4}{1.3} = \frac{0.1}{1.3} \approx 0.077[/tex]

A z-score of -0.69 corresponds to approximately 24.51% and a z-score of 0.077 corresponds to approximately 53.07%.

Therefore, the probability that it opens at temperatures between 19.5°C and 20.5°C is:

(
53.07% - 24.51% = 28.56%
)

Hence, there is a 28.56% probability of the thermostat operating between those temperatures.