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solve question 6-11 6 Sacha is shooting at a target which she has a probability ...

Jun 19, 2024

solve question 6-11

Solution by Steps

step 1

We need to find the probability that the proportion of times Sacha hits the target in her next 100 attempts is less than 0.8

step 2

Let

X

be the number of hits.

X

follows a binomial distribution with parameters

n = 100

and

p = 0.6

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 60

and

np(1-p) = 24

step 4

We standardize the variable:

Z = \frac{X - np}{\sqrt{np(1-p)}}

step 5

We need to find P\left(\frac{X}{100} < 0.8\right) = P\left(X < 80\right)

step 6

Standardizing: P\left(Z < \frac{80 - 60}{\sqrt{24}}\right) = P\left(Z < \frac{20}{\sqrt{24}}\right) = P\left(Z < 4.08\right)

step 7

Using the standard normal distribution table, P(Z < 4.08) \approx 1

# Part (b)

step 1

We need to find the probability that the proportion of times Sacha hits the target in her next 100 attempts is between 0.6 and 0.8

step 2

We need to find P(60 < X < 80)

step 3

Standardizing: P\left(\frac{60 - 60}{\sqrt{24}} < Z < \frac{80 - 60}{\sqrt{24}}\right) = P(0 < Z < 4.08)

step 4

Using the standard normal distribution table, P(0 < Z < 4.08) \approx 0.5

# Part (c)

step 1

We need to find the probability that the proportion of times Sacha hits the target in her next 100 attempts is between 0.7 and 0.8, given that it is more than 0.6

step 2

We need to find P(70 < X < 80 | X > 60)

step 3

Using conditional probability: P(70 < X < 80 | X > 60) = \frac{P(70 < X < 80)}{P(X > 60)}

step 4

We already know P(70 < X < 80) \approx 0.5 and P(X > 60) \approx 0.5

step 5

Therefore, P(70 < X < 80 | X > 60) = \frac{0.5}{0.5} = 1

6 Answer

(a) 1, (b) 0.5, (c) 1

Question 7 # Part (a)

step 1

We need to find the probability that the proportion of heads in the next 100 tosses of a fair coin is between 0.4 and 0.6

step 2

Let

X

be the number of heads.

X

follows a binomial distribution with parameters

n = 100

and

p = 0.5

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 50

and

np(1-p) = 25

step 4

We need to find P(40 < X < 60)

step 5

Standardizing: P\left(\frac{40 - 50}{\sqrt{25}} < Z < \frac{60 - 50}{\sqrt{25}}\right) = P(-2 < Z < 2)

step 6

Using the standard normal distribution table, P(-2 < Z < 2) \approx 0.9545

# Part (b)

step 1

We need to find the number of tosses

n

such that the probability that the proportion of heads is more than 0.55 is equal to 0.1

step 2

Let

X

be the number of heads.

X

follows a binomial distribution with parameters

n

and

p = 0.5

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 0.5n

and

np(1-p) = 0.25n

step 4

We need to find

n

such that P\left(\frac{X}{n} > 0.55\right) = 0.1

step 5

Standardizing: P\left(Z > \frac{0.55n - 0.5n}{\sqrt{0.25n}}\right) = 0.1

step 6

This simplifies to P\left(Z > \frac{0.05n}{\sqrt{0.25n}}\right) = 0.1

step 7

Solving for

n

: P\left(Z > \frac{0.05\sqrt{n}}{0.5}\right) = 0.1

step 8

Using the standard normal distribution table, P(Z > 1.28) = 0.1

step 9

Therefore,

\frac{0.05\sqrt{n}}{0.5} = 1.28 \Rightarrow \sqrt{n} = 25.6 \Rightarrow n = 655.36 \approx 655

7 Answer

(a) 0.9545, (b) 655

Question 8 # Part (a)

step 1

We need to find the probability that the proportion of defective items in the next batch of 1000 items produced is between 0.08 and 0.12

step 2

Let

X

be the number of defective items.

X

follows a binomial distribution with parameters

n = 1000

and

p = 0.1

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 100

and

np(1-p) = 90

step 4

We need to find P(80 < X < 120)

step 5

Standardizing: P\left(\frac{80 - 100}{\sqrt{90}} < Z < \frac{120 - 100}{\sqrt{90}}\right) = P(-2.11 < Z < 2.11)

step 6

Using the standard normal distribution table, P(-2.11 < Z < 2.11) \approx 0.9654

# Part (b)

step 1

We need to find the probability that the proportion of defective items in the next batch of 1000 items produced is between 0.08 and 0.12, given that it is greater than 0.10

step 2

We need to find P(80 < X < 120 | X > 100)

step 3

Using conditional probability: P(80 < X < 120 | X > 100) = \frac{P(80 < X < 120)}{P(X > 100)}

step 4

We already know P(80 < X < 120) \approx 0.9654 and P(X > 100) \approx 0.5

step 5

Therefore, P(80 < X < 120 | X > 100) = \frac{0.9654}{0.5} = 1.9308

8 Answer

(a) 0.9654, (b) 1.9308

Question 9 # Part (a)

step 1

We need to find the value of the sample proportion

\hat{p}

step 2

The sample proportion

\hat{p}

is given by

\hat{p} = \frac{x}{n}

, where

x

is the number of successes and

n

is the sample size

step 3

Here,

x = 230

and

n = 400

step 4

Therefore,

\hat{p} = \frac{230}{400} = 0.575

# Part (b)

step 1

We need to find the approximate probability that, in a random sample of 400 voters, the proportion who favour Candidate A is greater than or equal to this value of

\hat{p}

step 2

Let

X

be the number of voters who favour Candidate A.

X

follows a binomial distribution with parameters

n = 400

and

p = 0.52

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 208

and

np(1-p) = 99.84

step 4

We need to find

P(X \geq 230)

step 5

Standardizing:

P\left(Z \geq \frac{230 - 208}{\sqrt{99.84}}\right) = P\left(Z \geq 2.20\right)

step 6

Using the standard normal distribution table,

P(Z \geq 2.20) \approx 0.0139

9 Answer

(a) 0.575, (b) 0.0139

Question 10 # Part (a)

step 1

We need to find the value of the sample proportion

\hat{p}

step 2

The sample proportion

\hat{p}

is given by

\hat{p} = \frac{x}{n}

, where

x

is the number of successes and

n

is the sample size

step 3

Here,

x = 212

and

n = 250

step 4

Therefore,

\hat{p} = \frac{212}{250} = 0.848

# Part (b)

step 1

We need to find the approximate probability that, in a random sample of 250 batteries, the proportion lasting more than 100 hours is less than or equal to this value of

\hat{p}

step 2

Let

X

be the number of batteries lasting more than 100 hours.

X

follows a binomial distribution with parameters

n = 250

and

p = 0.9

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 225

and

np(1-p) = 22.5

step 4

We need to find

P(X \leq 212)

step 5

Standardizing:

P\left(Z \leq \frac{212 - 225}{\sqrt{22.5}}\right) = P\left(Z \leq -2.73\right)

step 6

Using the standard normal distribution table,

P(Z \leq -2.73) \approx 0.0032

# Part (c)

step 1

We need to determine if the probability found in part (b) causes us to doubt the manufacturer's claim

step 2

The probability

0.0032

is very small, indicating that it is highly unlikely for the sample proportion to be this low if the manufacturer's claim is true

step 3

Therefore, this result does cause us to doubt the manufacturer's claim

10 Answer

(a) 0.848, (b) 0.0032, (c) Yes

Question 11

step 1

We need to find the sample size

n

such that the probability that the proportion of people with wavy hair is less than 32% is equal to 0.2445

step 2

Let

X

be the number of people with wavy hair.

X

follows a binomial distribution with parameters

n

and

p = 0.35

step 3

We approximate the binomial distribution with a normal distribution:

X \sim N(np, np(1-p))

. Here,

np = 0.35n

and

np(1-p) = 0.2275n

step 4

We need to find

n

such that P\left(\frac{X}{n} < 0.32\right) = 0.2445

step 5

Standardizing: P\left(Z < \frac{0.32n - 0.35n}{\sqrt{0.2275n}}\right) = 0.2445

step 6

This simplifies to P\left(Z < \frac{-0.03n}{\sqrt{0.2275n}}\right) = 0.2445

step 7

Using the standard normal distribution table, P(Z < -0.69) = 0.2445

step 8

Therefore,

\frac{-0.03\sqrt{n}}{\sqrt{0.2275}} = -0.69 \Rightarrow \sqrt{n} = 5.25 \Rightarrow n = 27.56 \approx 28

11 Answer

Key Concept

Normal approximation to the binomial distribution

Explanation

The normal approximation to the binomial distribution is used when the sample size is large and the probability of success is not too close to 0 or 1. This allows us to use the properties of the normal distribution to approximate probabilities for binomially distributed variables.