The statistical qualty control department of Chips Ahoy Cookies had determined t

Question

The statistical qualty control department of Chips Ahoy Cookies had determined that the number of chocolate chips in an 18-ounce bag of cookies is approximately nommally distributed wilth a mean of 1252 chocalate chips and a standard deviation of 129 chocolate chips #44. Determine the probability that a randomly selected bag of cookies contains between 1000 and 1500 chocolate chips, inclusive with reference to problem #44 cookies contains fewer than 1100 chacolate chips. #45, determine the probability that a randomly selected bag of With reference to problem 044, determine the probability that a randomly selected bag of cookies contains more than 1200 chooolate chips #46 24

Explanation / Answer

Given: µ = 1252, = 129

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ [/n]. Since n = 1, Z = (X - µ)/

(a) For P (1000 < X < 1500) = P(X < 1500) – P(X < 1000)

For P( X < 1500)

Z = (1500 – 1252)/129 = 1.922

The probability for P(X < 1500) from the normal distribution tables is = 0.9727

For P( X < 1000)

Z = (1000 – 1252)/129 = -1.953

The probability for P(X < 1000) from the normal distribution tables is = 0.0254

Therefore the required probability is 0.9727 – 0.0254 = 0.9473

(b) For P( X < 1100)

Z = (1100 – 1252)/129 = -1.178

The required probability from the normal distribution tables is = 0.1193

(c) For P (X > 1200) = 1 - P (X < 1200), as the normal tables give us the left tailed probability only.

For P( X < 1200)

Z = (1200 – 1252)/129 = -0.403

The probability for P(X < 1200) from the normal distribution tables is = 0.3434

Therefore the required probability = 1 – 0.3434 = 0.6566

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