Question

1.)

For normally distributed data, what proportion of observations have z-scores satisfying the following conditions:

-2.27 < z < -1.89

Round to 4 decimal places.

2.)A particular fruit's weights are normally distributed, with a mean of 721 grams and a standard deviation of 32 grams.

The heaviest 16% of fruits weigh more than how many grams?

Give your answer to the nearest gram.

3.)

A particular fruit's weights are normally distributed, with a mean of 759 grams and a standard deviation of 38 grams.

If you pick 14 fruit at random, what is the probability that their mean weight will be between 726 grams and 743 grams.

Round to 4 decimal places.

Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

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4.)In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $46 and standard deviation of $2. Find the margin of error for a 95% confidence level.

5.)Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 288 with 47.9% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Solution :

1.) P( -2.27 < z < -1.89 )

= P(z < -1.89) - P(z < -2.27)

Using z table,

= 0.0294 - 0.0116

= 0.0178

Answer : Proportion = 0.0178

2.) mean = = 721

standard deviation = = 32

The z-distribution of the 16% is,

P(Z > z) = 16%

= 1 - P(Z < z ) = 0.16

= P(Z < ) = 1 - 0.16  

= P(Z < z ) = 0.84

= P(Z < 0.994 ) = 0.84

z = 0.994

Using z-score formula,

x = z * +

x = 0.994 * 32 + 721

x = 752.81

x = 753

Answer : 753 grams.

3.) mean = = 759

standard deviation = = 38

n = 14

= 759

= / n = 38 / 14 = 10.1559

P(726 < < 743)  

= P[(726 - 759) / 10.1559 < ( - ) / < (743 - 759) / 10.1559 )]

= P ( -3.25 < Z < -1.58)

= P(Z < -1.58 ) - P(Z < -3.25 )

Using z table,  

= 0.0571 - 0.0006

= 0.0565

Answer : Probability = 0.0565

4.) Point estimate = sample mean = = 46

Population standard deviation =    = 2

Sample size = n = 18

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z/2 * ( /n)

= 1.960 * ( 2 /  18 )

= 0.92

Answer : Margin of error = 0.92

5.) n = 288

Point estimate = sample proportion = = 47.9% = 0.479

1 - = 1 - 0.479 = 0.521

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

Z/2 = Z0.10 = 1.282

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.282 (((0.479 * 0.521) / 288)

= 0.038

A 80% confidence interval for population proportion p is ,

- E < p < + E

0.479 - 0.038 < p < 0.479 + 0.038

0.441 < p < 0.517

( 0.441 , 0.517 )

The 80% confidence interval for the population proportion p is : ( 0.441 , 0.517 )