Write an equation about having 6 puzzles, getting 5 more and having a lot now.

State the following numbers correct to the decimal place given 78.333 (2d.p) 0.06045 (4d.p) 288.598 (2d.p) 0.1008 (3d.p)

Alex Ar [27]

78.33 (3 rounds down)

0.0605 (5 rounds up)

288.60 (8 rounds up, but there is a nine, so nine becomes 0 and 5 becomes 6)

0.101 (8 rounds up)

6 0

3 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

A group of Construction students must choose their specialist options from the following list: Bricklaying, Damp-proofing, Drain

serious [3.7K]

Answer:

9

Step-by-step explanation:

Since drainage is to be included in the possible combinations, we can therefore determined the number from the remaining 6 options which can be in combination with drainage

(6 2) = 15 pairs.

Now there are restrictions placed, where we have no pairing of damp roofing with bricklaying or plastering. This removes two pairs, so we are left with 13pairs.

Another restrictions is that students choosing Joinery must also choose Flooring. This gives the three options: joinery, flooring and drained with no pairing with the remaining four pairs this removing another four pair.

So we are left with 9 pairs in total.

7 0

3 years ago

How 2 find intercept

Zarrin [17]

You can look at the graph and see where the point crosses the y - axis for the y-int and for the x-int you look to see where the point crosses the x-axis. The coordinate with always have a coordinate of zero.

8 0

3 years ago

20 POINTS PLS HELP ASP

svet-max [94.6K]

Answer:

Im thinking it is 50

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers