Question

**can you explain how to solve on calculator**

1. Assume that adults have IQ scores that are normally distributed with a mean of 96.9 and a standard deviation of 19.9.

Find the probability that a randomly selected adult has an IQ greater than 136.4

2. Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z= -.85, z= 1.26

3. Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. x=96

4. Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=20. Find the probability that a randomly selected adult has an IQ between 91 and 119.

5. Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=20.Find the probability that a randomly selected adult has an IQ less than 133.

6. Assume that females have pulse rates that are normally distributed with a mean of 75.0 beats per minute and a standard deviation of 12.5 beats per minute.

6a) If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

6b) If 16 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 69 beats per minute and 81 beats per minute.

6c) Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

7. Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z= -.81

8. The standard deviation of the distribution of sample means is _____

Answer 1

1.

µ = 96.9, σ = 19.9

P(X > 136.4) =

= P( (X-µ)/σ > (136.4-96.9)/19.9)

= P(z > 1.9849)

= 1 - P(z < 1.9849)

Using excel function:

= 1 - NORM.S.DIST(1.9849, 1)

= 0.0236

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2.

P(z < -0.85) =

Using excel function:

= NORM.S.DIST(-0.85, 1)

= 0.1977

P(z < 1.26) =

Using excel function:

= NORM.S.DIST(1.26, 1)

= 0.8962

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3.

µ = 100, σ = 15

P(X = 96) =

= P( (X-µ)/σ = (96-100)/15 )

= P(z = -0.2667)

Using excel function:

= NORM.S.DIST(-0.2667, 1)

= 0.3949

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4.

µ = 105, σ = 20

P(91 < X < 119) =

= P( (91-105)/20 < (X-µ)/σ < (119-105)/20 )

= P(-0.7 < z < 0.7)

= P(z < 0.7) - P(z < -0.7)

Using excel function:

= NORM.S.DIST(0.7, 1) - NORM.S.DIST(-0.7, 1)

= 0.5161

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5.

µ = 105, σ = 20

P(X < 133) =

= P( (X-µ)/σ < (133-105)/20 )

= P(z < 1.4)

Using excel function:

= NORM.S.DIST(1.4, 1)

= 0.9192

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6. a)

P(69 < X < 81) =

= P( (69-75)/12.5 < (X-µ)/σ < (81-75)/12.5 )

= P(-0.48 < z < 0.48)

= P(z < 0.48) - P(z < -0.48)

Using excel function:

= NORM.S.DIST(0.48, 1) - NORM.S.DIST(-0.48, 1)

= 0.3688

b)

µ = 75, σ = 12.5, n = 16

P(69 < X̅ < 81) =

= P( (69-75)/(12.5/√16) < (X-µ)/(σ/√n) < (81-75)/(12.5/√16) )

= P(-1.92 < z < 1.92)

= P(z < 1.92) - P(z < -1.92)

Using excel function:

= NORM.S.DIST(1.92, 1) - NORM.S.DIST(-1.92, 1)

= 0.9451

c) Because the population is normally distributed so the sample is also normally distributed.

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7.

P(z < -0.81) =

Using excel function:

= NORM.S.DIST(-0.81, 1)

= 0.209