Question

1. A manufacturer knows that their items have a normally distributed lifespan, with a mean of 14.1 years, and standard deviation of 2.8 years.

If you randomly purchase one item, what is the probability it will last longer than 20 years?

2.

A particular fruit's weights are normally distributed, with a mean of 506 grams and a standard deviation of 21 grams.

If you pick one fruit at random, what is the probability that it will weigh between 475 grams and 565 grams

3.

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.2 inches, and standard deviation of 6.1 inches.

A) What is the probability that a randomly chosen child has a height of less than 69.75 inches?

Answer= (Round your answer to 3 decimal places.)

B) What is the probability that a randomly chosen child has a height of more than 37.1 inches?

Answer= (Round your answer to 3 decimal places.)

4.

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.9 inches, and standard deviation of 3.4 inches.

What is the probability that the height of a randomly chosen child is between 61.1 and 63.7 inches? Do not round until you get your your final answer, and then round to 3 decimal places.

Answer= (Round your answer to 3 decimal places.)

Answer 1

where -> Mean

-> Standard deviation

1) = 14.1 years

= 2.8 years

Probability that it will last longer than 20 years

= P(X > 20) = P{Z > (20 - 14.1)/2.8}

= P(Z > 2.107)

= 0.0176

2) = 506 grams

= 21 grams

The required probability = P(475 < X < 565)

= P{(475 - 506)/21 < Z < (565 - 506)/21}

= P(-1.476 < Z < 2.81)

= 0.9275

3) = 55.2 inches

= 6.1 inches

A) The required probability = P(X < 69.75)

= P{Z < (69.75 - 55.2)/6.1}

= P(Z < 2.385)

= 0.9914

B) The required probability = P(X > 37.1)

= P{Z > (37.1 - 55.2)/6.1}

= P(Z > -2.967)

= 0.9985

4) = 55.9 inches

= 3.4 inches

The required probability = P(61.1 < X < 63.7)

= P(1.53 < Z < 2.29)

= 0.052