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borishaifa [10]
3 years ago
10

A box contains recipes from five categories. The following table shows the possible categories and the associated probability fo

r a recipe randomly chosen.
Genre Probability
Main Dish 0.421
Appetizer 0.210
Desert 0.103
Salad 0.173
Vegetable 0.093
1. What is the probability of randomly selecting anything but a vegetable or a main dish category?
Mathematics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

0.486

Step-by-step explanation:

Given that a box contains recipes from five categories.

The following table shows the possible categories and the associated probability for a recipe randomly chosen.

Genre Probability

Main Dish 0.421

Appetizer 0.210

Desert 0.103

Salad 0.173

Vegetable 0.093

This is a proper probability distribution function because total prob =1 and also each prob lies between 0 and 1.

The  probability of randomly selecting anything but a vegetable or a main dish category

= Prob of selecting appetizer or dessert or salad

= 0.210+0.103+0.173\\=0.486

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The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

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Standard deviation, σ = 65

Sample size, n = 150

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Level of significance = 0.01

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$H_1 : \mu > 1900$

Test statics :

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$z=0.38$

Finding the p-value:

P-value = P(Z > z)

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From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

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Decision :

If the p value is less than 0.01, then we reject the H_0, otherwise we fail to reject  H_0.

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  H_0.

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

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