Mathematics, 29.07.2020 02:01 mrstealyogirl40

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92. A. If a random sample of 25 specimens is selected, what is theprobability that the sample average sediment density is at most3.00? Between 2.68 and 3.00
B. How large a sample size would be required to ensure thatthe first probability in part (a) is at least .99 ?

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Answer from: ashley24109

The sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.92.

Let $\bar X$ = sample average sediment density

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.68

$\sigma$ = population standard deviation = 0.92

n = sample of specimens = 25

(a) The probability that the sample average sediment density is at most 3.00 is given by = P( $\bar X$ $\leq$ 3.00)

P( $\bar X$ $\leq$ 3.00) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z $\leq$ 1.74) = 0.9591

The above probability is calculated by looking at the value of x = 1.74 in the z table which has an area of 0.9591.

Also, the probability that the sample average sediment density is between 2.68 and 3.00 is given by = P(2.68 < $\bar X$ < 3.00)

P(2.68 < $\bar X$ < 3.00) = P( $\bar X$ < 3.00) - P( $\bar X$ $\leq$ 2.68)

P( $\bar X$ < 3.00) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3.00-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z < 1.74) = 0.9591

P( $\bar X$ $\leq$ 2.68) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{2.68-2.68}{\frac{0.92}{\sqrt{25} } }$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.74 and x = 0 in the z table which has an area of 0.9591 and 0.50.

Therefore, P(2.68 < $\bar X$ < 3.00) = 0.9591 - 0.50 = 0.4591.

(b) Now, we have to find a sample size that would ensure that the first probability in part (a) is at least 0.99, that is;

P( $\bar X$ $\leq$ 3.00) $\geq$ 0.99

P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }$ ) $\geq$ 0.99

P(Z $\leq$ $\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }$ ) $\geq$ 0.99

Now, in the z table; the critical value of x which has an area of at least 0.99 is given by 2.3263, that is;

$\frac{3.00-2.68}{\frac{0.92}{\sqrt{n} } }=2.3263$

$\sqrt{n} } }=\frac{ 2.3263\times 0.92}{0.32}$

$\sqrt{n} } }=6.69$

n = 44.76 ≈ 45 {By squaring both sides}

Hence, the sample size must be 45 large enough that would ensure that the first probability in part (a) is at least 0.99.

Answer from: Quest

38+105= 143, 39+104=143, and you will find then adding the numbers on opposite sides, moving inwards yields the answer 138 every time. because of this, you’re going to multiply 138 times the number of pairs, which is 34. therefore, 34*138= 4692(i’m sorry if it’s the wrong numbers, but it’s the right concept)
I’m not sure how to solve these type of problems