Label each part of the equation. Write operation or variable N-100+r=54

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the act

stealth61 [152]

Answer: 0.2551

Step-by-step explanation:

Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the actual temperature of the medium, and standard deviation σ.

Significance level : $\alpha=1-0.95=0.05$

The critical z-value for 95% confidence : $z_{\alpha/2}=1.960$ (1)

Since , $z=\dfrac{x-\mu}{\sigma}$ (where x be any random variable that represents the temperature reading from a thermocouple.)

Then, from (1)

$\dfrac{x-\mu}{\sigma}=1.96\\\\ x-\mu=1.96\sigma$ (2)

Also, all readings are within 0.5° of μ,

i.e. $x-\mu$

i.e. $1.96\sigma$ [From (2)]

i.e. $\sigma$

i.e. $\sigma\approx0.2551$

The required standard deviation : $\sigma=0.2551$

3 0

2 years ago

g Assume that the distribution of time spent on leisure activities by adults living in household with no young children is norma

OLga [1]

Answer:

"The probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day" is about 0.8749.

Step-by-step explanation:

We have here a random variable that is normally distributed, namely, the time spent on leisure activities by adults living in a household with no young children.

The normal distribution is determined by two parameters: the population mean, $\\ \mu$ , and the population standard deviation, $\\ \sigma$ . In this case, the variable follows a normal distribution with parameters $\\ \mu = 4.5$ hours per day and $\\ \sigma = 1.3$ hours per day.

We can solve this question following the next strategy:

Use the cumulative standard normal distribution to find the probability.
Find the z-score for the raw score given in the question, that is, x = 6 hours per day.
With the z-score at hand, we can find this probability using a table with the values for the cumulative standard normal distribution. This table is called the standard normal table, and it is available on the Internet or in any Statistics books. Of course, we can also find these probabilities using statistics software or spreadsheets.

We use the standard normal distribution because we can "transform" any raw score into standardized values, which represent distances from the population mean in standard deviations units, where a positive value indicates that the value is above the mean and a negative value that the value is below it. A standard normal distribution has $\\ \mu = 0$ and $\\ \sigma = 1$ .

The formula for the z-scores is as follows

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Solving the question

Using all the previous information and using formula [1], we have

x = 6 hours per day (the raw score).

$\\ \mu = 4.5$ hours per day.

$\\ \sigma = 1.3$ hours per day.

Then (without using units)

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{6 - 4.5}{1.3}$

$\\ z = \frac{1.5}{1.3}$

$\\ z = 1.15384 \approx 1.15$

We round the value of z to two decimals since most standard normal tables only have two decimals for z.

We can observe that z = 1.15, and it tells us that the value is 1.15 standard deviations units above the mean.

With this value for z, we can consult the cumulative standard normal table, and for this z = 1.15, we have a cumulative probability of 0.8749. That is, this table gives us P(z<1.15).

We can describe the procedure of finding this probability in the next way: At the left of the table, we have z = 1.1; we can follow the first line on the table until we find 0.05. With these two values, we can determine the probability obtained above, P(z<1.15) = 0.8749.

Notice that the probability for the z-score, P(z<1.15), of the raw score, P(x<6) are practically the same, $\\ P(z$ . For an exact probability, we have to use a z-score = 1.15384 (without rounding), that is, $\\ P(z$ . However, the probability is approximated since we have to round z = 1.15384 to z = 1.15 because of the use of the table.

Therefore, "the probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day" is about 0.8749.

We can see this result in the graphs below. First, for P(x<6) in $\\ N(4.5, 1.3)$ (red area), and second, using the standard normal distribution ( $\\ N(0, 1)$ ), for P(z<1.15), which corresponds with the blue shaded area.