, Stevens glass holds 40 milliliters of milk fairs holds 2/5 as much milk. How much milk does fairs hold

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°

Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 98.6, \sigma = 0.6, n = 36, s = \frac{0.6}{\sqrt{36}} = 0.1$

If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4degrees° F.

This is 1 subtracted by the pvalue of Z when X = 98.4. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{98.4 - 98.6}{0.1}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

6 0

2 years ago

A rectangle has a length at most two more than seven times the width. If the perimeter is less than or equal to 52 units, find t

ella [17]

$7x + 2 \leqslant 52$

7 0

2 years ago

Petersons bought 249,000 townhome. They made a down payment of $42000.00 and took mortgage for the rest. Over the course of 30 y

xenn [34]

Answer:

Answer in photo

6 0

2 years ago

A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean

elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}$ ). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0

2 years ago

Find two numbers with a sum of 20 and a difference of 14.

vivado [14]

A and B Definitely. 3 and 17, -3 and -17

4 0

2 years ago