Mathematics, 30.06.2020 17:01 aylineorozco836

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm. For a randomly found shard, find the following probabilities. a. the thickness is less than 3.0 mm
b. the thickness is more than 7.0 mm
c. the thickness is between 3.0 mm and 7.0 mm

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Answer from: genyjoannerubiera

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.

So, X ~ Normal( $\mu=5.0,\sigma^{2} =1.7^{2}$ )

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

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

where, $\mu$ = population mean thickness = 5.0 mm

$\sigma$ = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.0-5.0}{1.7}$ ) = P(Z < -1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{7.0-5.0}{1.7}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X $\leq$ 3.0 mm)

P(X < 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{7.0-5.0}{1.7}$ ) = P(Z < 1.18) = 0.881

P(X $\leq$ 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.0-5.0}{1.7}$ ) = P(Z $\leq$ -1.18) = 1 - P(Z < 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

Answer from: Quest

what are the options? ?

step-by-step explanation:

Point e is the midpoint of and point f is the midpoint of . which statements about the figure must

Answer from: Quest

1.17+3.99a. you combine the terms with a to create 3.99 and the 1.17 stays because it has no like terms or a variable

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